Genetics - Analysis of Genes and Genomes, 9th edition

Citation preview

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved. Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 18:44:16.

NINTH EDITION

GENETICS ANALYSIS OF GENES AND GENOMES

Daniel L. Hartl, PhD Harvard University Cambridge, Massachusetts

Bruce J. Cochrane, PhD

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Miami University Oxford, Ohio

JONES & BARTLETT LEARNING

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 18:44:47.

World Headquarters Jones & Bartlett Learning 5 Wall Street Burlington, MA 01803 978-443-5000 [email protected] www.jblearning.com Jones & Bartlett Learning books and products are available through most bookstores and online booksellers. To contact Jones & Bartlett Learning directly, call 800-832-0034, fax 978-443-8000, or visit our website, www.jblearning.com.

Substantial discounts on bulk quantities of Jones & Bartlett Learning publications are available to corporations, professional associations, and other qualified organizations. For details and specific discount information, contact the special sales department at Jones & Bartlett Learning via the above contact information or send an email to [email protected].

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Copyright © 2019 by Jones & Bartlett Learning, LLC, an Ascend Learning Company All rights reserved. No part of the material protected by this copyright may be reproduced or utilized in any form, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the copyright owner. The content, statements, views, and opinions herein are the sole expression of the respective authors and not that of Jones &

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 18:45:06.

Bartlett Learning, LLC. Reference herein to any specific commercial product, process, or service by trade name, trademark, manufacturer, or otherwise does not constitute or imply its endorsement or recommendation by Jones & Bartlett Learning, LLC and such reference shall not be used for advertising or product endorsement purposes. All trademarks displayed are the trademarks of the parties noted herein. Genetics: Analysis of Genes and Genomes, Ninth Edition is an independent publication and has not been authorized, sponsored, or otherwise approved by the owners of the trademarks or service marks referenced in this product. There may be images in this book that feature models; these models do not necessarily endorse, represent, or participate in the activities represented in the images. Any screenshots in this product are for educational and instructive purposes only. Any individuals and scenarios featured in the case studies throughout this product may be real or fictitious, but are used for instructional purposes only.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

13660-9 Production Credits VP, Product Management: David D. Cella Director of Product Management: Matthew Kane Product Specialist: Audrey Schwinn Product Assistant: Loren-Marie Durr Senior Production Editor: Nancy Hitchcock Marketing Manager: Lindsay White Production Services Manager: Colleen Lamy Manufacturing and Inventory Control Supervisor: Amy Bacus Composition: SourceHOV LLC Cover Design: Scott Moden

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 18:45:06.

Rights & Media Specialist: Wes DeShano Media Development Editor: Troy Liston Cover Image: Getty Images/Science Photo Library RF Printing and Binding: LSC Communications Cover Printing: LSC Communications Library of Congress Cataloging-in-Publication Data Names: Hartl, Daniel L., author. | Cochrane, Bruce, author. Title: Genetics : analysis of genes and genomes / Daniel L. Hartl & Bruce Cochrane. Description: Ninth edition. | Burlington, Massachusetts : Jones & Bartlett Learning, [2018] | Includes bibliographical references and index. Identifiers: LCCN 2017030082 | ISBN 9781284122930 Subjects: | MESH: Genetic Phenomena | Genetic Techniques | Genetic Structures | Genomics Classification: LCC QH430 | NLM QU 500 | DDC 576.5--dc23 LC record available at https://secureweb.cisco.com/2017030082 6048

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Printed in the United States of America 21 20 19 18 17 10 9 8 7 6 5 4 3 2 1

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 18:45:06.

Brief Contents Unit I Defining and Working with GENES

1 Genes, Genomes, and Genetic Analysis 2 DNA Structure and Genetic Variation Unit II Transmission Genetics

3 Mendelian Genetics: The Principles of Segregation and Assortment

4 The Chromosomal Basis of Inheritance 5 Genetic Linkage and Chromosome Mapping 6 Human Karyotypes and Chromosome Behavior 7 The Genetic Basis of Complex Traits 8 Genetics of Bacteria and Their Viruses

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Unit III Organization and Replication of Chromosomes and DNA

9 Molecular Organization of Chromosomes and Genomes

10 DNA Replication and Sequencing 11 Mutation, Repair, and Recombination Unit IV Gene Expression

12 Molecular Biology of Gene Expression 13 Molecular Mechanisms of Gene Regulation

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:07:16.

14 Manipulating Genes and Genomes 15 Genetic Control of Development 16 Molecular Genetics of the Cell Cycle and Cancer Unit V Variation

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

17 Mitochondrial DNA and Extranuclear Inheritance 18 Genes in Populations 19 Molecular and Human Evolutionary Genetics

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:07:16.

Contents Preface The Student Experience What's New in the Ninth Edition Teaching Tools Acknowledgements About the Authors

Unit I Defining and Working with GENES

1 Genes, Genomes, and Genetic Analysis Learning Objectives & Science Competencies 1.1 DNA as the Genetic Material Experimental Proof of the Genetic Function of DNA Genetic Role of DNA in Bacteriophages 1.2 DNA Structure and Replication

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

An Overview of DNA Replication 1.3 Genes and Proteins Inborn Errors of Metabolism as a Cause of Hereditary Disease 1.4 Genetic Analysis Mutant Genes and Defective Proteins Complementation Test for Mutations in the Same Gene Analysis of Complementation Data Other Applications of Genetic Analysis 1.5 Gene Expression: The Central Dogma Transcription

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Translation The Genetic Code 1.6 Mutation and Variation Variation in Populations 1.7 Genes and Environment 1.8 The Molecular Unity of Life Prokarya, Archaea, and Eukarya Genomes and Proteomes ROOTS OF DISCOVERY: The Black Urine Disease ROOTS OF DISCOVERY: One Gene, One Enzyme

2 DNA Structure and Genetic Variation Learning Objectives & Science Competencies 2.1 Genetic Differences Among Individuals DNA Markers as Landmarks in Chromosomes 2.2 The Terminology of Genetic Analysis

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.3 The Molecular Structure of DNA Polynucleotide Chains The Double Helix Base Pairing and Base Stacking Antiparallel Strands DNA Structure as Related to Function 2.4 The Separation and Identification of Genomic DNA Fragments Restriction Enzymes and Site-Specific DNA Cleavage Gel Electrophoresis Nucleic Acid Hybridization 2.5 Amplification of Specific DNA for Detection and Purification Constraints on DNA Replication: Primers and 5′-to-3′ Strand Elongation

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

The Polymerase Chain Reaction 2.6 Types of DNA Markers Present in Genomic DNA Restriction Fragment Length Polymorphisms Single-Nucleotide Polymorphisms Tandem Repeat Polymorphisms Copy-Number Variation 2.7 Applications of DNA Markers Genetic Markers, Genetic Mapping, and “Disease Genes” Other Uses for DNA Markers ROOTS OF DISCOVERY: The Double Helix THE CUTTING EDGE: High-Throughput SNP Genotyping

Unit II Transmission Genetics

3 Mendelian Genetics: The Principles of Segregation and Assortment Learning Objectives & Science Competencies 3.1 Morphological and Molecular Phenotypes 3.2 Segregation of a Single Gene

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Phenotypic Ratios in the F2 Generation The Principle of Segregation Verification of Segregation The Testcross and the Backcross 3.3 Segregation of Two or More Genes The Principle of Independent Assortment The Testcross with Independently Assorting Genes Three or More Genes 3.4 Human Pedigree Analysis Characteristics of Dominant and Recessive Inheritance Most Human Genetic Variation Is Not “Bad” Molecular Markers in Human Pedigrees

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

3.5 Incomplete Dominance and Epistasis Multiple Alleles Human ABO Blood Groups Epistasis 3.6 Probability in Genetic Analysis Elementary Outcomes and Events Probability of the Union of Events Probability of the Intersection of Events 3.7 Conditional Probability and Pedigrees Bayes' Theorem ROOTS OF DISCOVERY: What Did Gregor Mendel Think He Discovered? ROOTS OF DISCOVERY: This Land Is Your Land

4 The Chromosomal Basis of Inheritance Learning Objectives & Science Competencies 4.1 The Stability of Chromosome Complements 4.2 Mitosis

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Prophase Metaphase Anaphase Telophase 4.3 Meiosis The First Meiotic Division: Reduction The Second Meiotic Division: Equation 4.4 Sex-Chromosome Inheritance Chromosomal Determination of Sex X-Linked Inheritance Pedigree Characteristics of Human X-Linked Inheritance Heterogametic Females

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Nondisjunction as Proof of the Chromosome Theory of Heredity Sex Determination in Drosophila 4.5 Probability in the Prediction of Progeny Distributions Using the Binomial Distribution in Genetics Meaning of the Binomial Coefficient 4.6 Testing Goodness of Fit to a Genetic Hypothesis Random Variables and Distributions The Chi-Square Method Are Mendel's Data Too Good to Be True? ROOTS OF DISCOVERY: Grasshopper, Grasshopper ROOTS OF DISCOVERY: The White-Eyed Male

5 Genetic Linkage and Chromosome Mapping Learning Objectives & Science Competencies 5.1 Linkage and Recombination of Genes in a Chromosome

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Coupling Versus Repulsion of Syntenic Alleles The Chi-Square Test for Linkage Each Pair of Linked Genes Has a Characteristic Frequency of Recombination Recombination in Females Versus Males 5.2 Genetic Mapping Map Distance and Frequency of Recombination Crossing Over Recombination Between Genes Results from a Physical Exchange Between Chromosomes Crossing Over Takes Place at the Four-Strand Stage of Meiosis Multiple Crossovers 5.3 Genetic Mapping in a Three-Point Testcross Chromosome Interference in Double Crossovers

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Genetic Mapping Functions Genetic Map Distance and Physical Distance 5.4 Mapping by Tetrad Analysis Unordered Tetrads Ordered Tetrads 5.5 Genetic Mapping in Humans Inferring Linkage in Pedigrees Association Mapping 5.6 Special Features of Recombination Recombination Within Genes Mitotic Recombination ROOTS OF DISCOVERY: Genes All in a Row ROOTS OF DISCOVERY: Mapping Markers in the Human Genome

6 Human Karyotypes and Chromosome Behavior Learning Objectives & Science Competencies

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

6.1 The Human Karyotype Standard Karyotypes The Centromere and Chromosome Stability Dosage Compensation of X-Linked Genes The Calico Cat Pseudoautosomal Inheritance Active Genes in the “Inactive” X Chromosome Gene Content and Evolution of the Y Chromosome 6.2 Chromosome Abnormalities in Human Pregnancies Down Syndrome and Other Viable Trisomies Trisomic Segregation Sex-Chromosome Abnormalities Environmental Effects on Nondisjunction 6.3 Chromosomal Deletions and Duplications

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Deletions Deletion Mapping Duplications Unequal Crossing Over in Human Red–Green Color Blindness Copy-Number Variation with Reciprocal Risks of Autism and Schizophrenia Gene Duplication and the Evolution of New Proteins 6.4 Genetics of Chromosomal Inversions Paracentric Inversion (Not Including the Centromere) Pericentric Inversion (Including the Centromere) 6.5 Chromosomal Translocations Reciprocal Translocations Pseudolinkage in Heterozygotes Robertsonian Translocations and Trisomy 21 Translocation Complexes in Oenothera 6.6 Genomic Position Effects on Gene Expression 6.7 Polyploidy in Plant Evolution

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Sexual Versus Asexual Polyploidization Autopolyploids and Allopolyploids Monoploid Organisms ROOTS OF DISCOVERY: Lyonization of an X Chromosome ROOTS OF DISCOVERY: The First Human Chromosomal Disorder Identified

7 The Genetic Basis of Complex Traits Learning Objectives & Science Competencies 7.1 Complex Traits Continuous, Categorical, and Threshold Traits The Normal Distribution Reconciling Mendelian Inheritance with Continuous Traits

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

7.2 Causes of Variation Genotypic Variation Environmental Variation Genetics and Environment Combined Genotype-by-Environment Interaction and Association 7.3 Heritability The Number of Genes Affecting Complex Traits Broad-Sense Heritability Narrow-Sense Heritability Twin Studies 7.4 Correlation Between Relatives Covariance and Correlation The Graphical Meaning of a Correlation 7.5 Heritability and Selection Phenotypic Change with Individual Selection: A Prediction Equation Long-Term Artificial Selection Heritabilities of Threshold Traits 7.6 Misconceptions About Heritability

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

7.7 Identification of Genes Affecting Complex Traits Linkage Analysis in the Genetic Mapping of Quantitative-Trait Loci Genome-Wide Association Studies Candidate Genes for Complex Traits ROOTS OF DISCOVERY: A Maize'n Grass THE CUTTING EDGE: Crowd-Sourced Genomics

8 Genetics of Bacteria and Their Viruses Learning Objectives & Science Competencies 8.1 Plasmids and Genetic Exchange The F Plasmid: A Conjugative Plasmid

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

8.2 Bacterial Genetics Mutant Phenotypes Isolating Auxotrophs Mechanisms of Genetic Exchange 8.3 DNA-Mediated Transformation 8.4 Genetic Exchange and Conjugation Cointegrate Formation and Hfr Cells Time-of-Entry Mapping F′ Plasmids 8.5 Transduction The Phage Lytic Cycle Generalized Transduction 8.6 Bacteriophage Genetics Plaque Formation and Phage Mutants Genetic Recombination in the Lytic Cycle Fine Structure of the rII Gene in Bacteriophage T4 8.7 Lysogeny and Specialized Transduction Site-Specific Recombination and Lysogeny Specialized Transduction ROOTS OF DISCOVERY: The Sex Life of Bacteria ROOTS OF DISCOVERY: Artoo

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

THE CUTTING EDGE: Surviving in a Hostile World

Unit III Organization and Replication of Chromosomes and DNA

9 Molecular Organization of Chromosomes and Genomes Learning Objectives & Science Competencies 9.1 Genome Size and Evolutionary Complexity: The CValue Paradox 9.2 The Supercoiling of DNA

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Topoisomerase Enzymes 9.3 The Structure of Bacterial Genomes Mobile DNA in Prokaryotes Mobilization of Nonconjugative Plasmids 9.4 The Structure of Eukaryotic Genomes The Nucleosome: The Structural Unit of Chromatin The Nucleosome Core Particle Chromosome Territories in the Nucleus Chromosome Condensation Polytene Chromosomes 9.5 Analysis of Sequence Complexity in Eukaryotic Genomes

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Kinetics of DNA Renaturation Analysis of Genome Size and Repetitive Sequences by Renaturation Kinetics 9.6 Unique and Repetitive Sequences in Eukaryotic Genomes Unique Sequences Highly Repetitive Sequences Middle-Repetitive Sequences and Transposable Elements Molecular Mechanisms of Transposition Transposable Elements in the Human Genome 9.7 Molecular Structure of the Centromere 9.8 Molecular Structure of the Telomere Telomere Length Limits the Number of Cell Doublings ROOTS OF DISCOVERY: Telomeres: The Beginning of the End

10 DNA Replication and Sequencing Learning Objectives & Science Competencies

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

10.1 Problems of Initiation, Elongation, and Incorporation Error 10.2 Semiconservative Replication of Double-Stranded DNA The Meselson–Stahl Experiment Theta Replication of Circular DNA Molecules Rolling-Circle Replication 10.3 Initiation of DNA Replication Unwinding, Stabilization, and Stress Release Formation of the Primosome Complex 10.4 The Elongation Process and Proofreading Discontinuous Replication of the Lagging Strand The Joining of Precursor Fragments 10.5 Terminating Replication 10.6 Replication in Eukaryotes

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Semiconservative Replication of DNA in Chromosomes Multiple Origins and Bidirectional Replication in Eukaryotes Initiation Termination and the Problem of Telomeres 10.7 Exploiting the System: DNA Sequencing Sanger Sequencing Next-Generation Sequencing ROOTS OF DISCOVERY: Replication by Halves

11 Mutation, Repair, and Recombination Learning Objectives & Science Competencies 11.1 Types of Mutations Germ-Line and Somatic Mutations Conditional Mutations

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Classification by Function 11.2 The Molecular Basis of Mutation Nucleotide Substitutions Missense Mutations: The Example of Sickle-Cell Anemia Insertions, Deletions, and Frameshift Mutations Dynamic Mutation of Trinucleotide Repeats Cytosine Methylation and Gene Inactivation 11.3 Transposable Elements as Agents of Mutation 11.4 Spontaneous Mutation The Nonadaptive Nature of Mutation Estimation of Mutation Rates Hotspots of Mutation 11.5 Mutagens

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Depurination Oxidation Base-Analog Mutagens Chemical Agents That Modify DNA Ultraviolet Irradiation Ionizing Radiation Genetic Effects of the Chernobyl Nuclear Accident 11.6 Mechanisms of DNA Repair Mismatch Repair Base-Excision Repair AP Repair Nucleotide-Excision Repair Photoreactivation DNA Damage Bypass Double-Strand Gap Repair The SOS Repair System 11.7 Reverse Mutations and Suppressor Mutations Intragenic Suppression

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Intergenic Suppression The Ames Test for Mutagen/Carcinogen Detection 11.8 The Relationship Between Repair and Recombination Double-Strand Break and Repair Model Hotspots of Recombination ROOTS OF DISCOVERY: X-Ray Daze

Unit IV Gene Expression

12 Molecular Biology of Gene Expression Learning Objectives & Science Competencies 12.1 Amino Acids, Polypeptides, and Proteins 12.2 Colinearity Between Coding Sequences and Polypeptides 12.3 Overview of Transcription 12.4 Transcription in Prokaryotes

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Initiation Genetic Evidence for Promoters Elongation Termination 12.5 Transcription in Eukaryotes RNA Polymerases Promoters and Promoter Recognition Initiation Elongation Termination 12.6 Messenger RNA and RNA Processing 5′ Capping and 3′ Polyadenylation Splicing of Intervening Sequences Characteristics of Human Transcripts Coupling of Transcription and RNA Processing Mechanism of RNA Splicing

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Effects of Intron Mutations Exon Shuffling in the Origin of New Genes 12.7 Protein Synthesis Initiation in Prokaryotes Initiation in Eukaryotes Elongation Release Translational Proofreading and Premature Termination Nonsense-Mediated Decay Polysomes 12.8 Protein Folding and Chaperones 12.9 The Standard Genetic Code Genetic Evidence for a Triplet Code How the Code Was Cracked Features of the Standard Code Transfer RNA and Aminoacyl-tRNA Synthetase Enzymes Redundancy and Wobble Nonsense Suppression ROOTS OF DISCOVERY: Messenger “Light”

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

ROOTS OF DISCOVERY: Poly-U

13 Molecular Mechanisms of Gene Regulation Learning Objectives & Science Competencies 13.1 Transcriptional Regulation in Prokaryotes Inducible and Repressible Systems of Negative Regulation Positive Regulation Stochastic Noise in Gene Expression 13.2 The Operon System of Gene Regulation Lac− Mutants

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Inducible and Constitutive Synthesis and Repression Repressors, Operators, and Promoters The Operon System of Transcriptional Regulation Stochastic Noise in Lac Expression Positive Regulation of the Lactose Operon Regulation of the Tryptophan Operon 13.3 Regulation Through Transcription Termination Attenuation Riboswitches 13.4 Regulation in Bacteriophage Lambda 13.5 Transcriptional Regulation in Eukaryotes

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Galactose Metabolism in Yeast Transcriptional Activator Proteins Transcriptional Enhancers and Transcriptional Silencers Deletion Scanning The Eukaryotic Transcription Complex Chromatin-Remodeling Complexes Alternative Promoters 13.6 Epigenetic Mechanisms of Transcriptional Regulation Cytosine Methylation Methylation and Transcriptional Inactivation Genomic Imprinting in the Female and Male Germ Lines 13.7 Regulation Through RNA Processing and Decay Alternative Splicing Messenger RNA Stability 13.8 Noncoding RNAs and Regulation Interfering RNAs Long Noncoding RNA 13.9 Translational Control Regulatory RNAs Controlling Translation

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

13.10 Programmed DNA Rearrangements Gene Amplification Antibody and T-Cell Receptor Variability Mating-Type Interconversion Transcriptional Control of Mating Type ROOTS OF DISCOVERY: Operator? Operator? ROOTS OF DISCOVERY: Double Trouble

14 Manipulating Genes and Genomes Learning Objectives & Science Competencies 14.1 Site-Specific DNA Cleavage and Cloning Vectors Production of DNA Fragments with Defined Ends Recombinant DNA Molecules Plasmid, Lambda, and Cosmid Vectors 14.2 Cloning Strategies Joining DNA Fragments Insertion of a Particular DNA Molecule into a Vector The Use of Reverse Transcriptase: cDNA and RT-PCR 14.3 Detection of Recombinant Molecules

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Gene Inactivation in the Vector Molecule Screening for Particular Recombinants 14.4 Genomics Genomic Sequencing Sequencing the Human Genome Genome Annotation Comparative Genomics 14.5 Functional Genomics Chromatin Immunoprecipitation Two-Hybrid Analysis of Protein–Protein Interactions 14.6 Transgenic Organisms Germ-Line Transformation in Animals

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Genetic Engineering in Plants Transformation Rescue Site-Directed Mutagenesis and Knockout Mutations Gene Expression “Knockdown” with RNAi 14.7 Gene Editing CRISPR–Cas9 in Practice 14.8 Some Applications of Genetic Engineering Giant Salmon with Engineered Growth Hormone Nutritionally Engineered Rice Production of Useful Proteins Genetic Engineering with Animal Viruses ROOTS OF DISCOVERY: Hello, Dolly! THE CUTTING EDGE: Editing the Muscular Dystrophy Gene

15 Genetic Control of Development Learning Objectives & Science Competencies 15.1 Genetic Determinants of Development

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

15.2 Early Embryonic Development in Animals Autonomous Development and Intercellular Signaling Composition and Organization of Oocytes Early Development and Activation of the Zygotic Genome 15.3 Genetic Analysis of Development in the Nematode Analysis of Cell Lineages Mutations Affecting Cell Lineages Programmed Cell Death Loss-of-Function and Gain-of-Function Alleles Epistasis in the Analysis of Developmental Switches 15.4 Genetic Control of Development in Drosophila Maternal-Effect Genes and Zygotic Genes

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Genetic Basis of Pattern Formation in Early Development Coordinate Genes Gap Genes Pair-Rule Genes Segment-Polarity Genes Interactions in the Regulatory Hierarchy Metamorphosis of the Adult Fly Homeotic Genes Master Control Genes in Evolution 15.5 Regulatory RNAs in Development Micro RNA–Based Regulation in C. elegans miRNAs in HOX Gene Regulation Long Intergenic Noncoding RNAs 15.6 Genetic Control of Development in Higher Plants Flower Development in Arabidopsis Combinatorial Determination of the Floral Organs ROOTS OF DISCOVERY: Distinguished Lineages ROOTS OF DISCOVERY: Embryogenesis

16 Molecular Genetics of the Cell Cycle and Cancer Learning Objectives & Science Competencies Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

16.1 The Cell Cycle Key Events in the Cell Cycle Transcriptional Program of the Cell Cycle 16.2 Genetic Analysis of the Cell Cycle Mutations Affecting Progression Through the Cell Cycle 16.3 Progression Through the Cell Cycle Cyclins and Cyclin-Dependent Protein Kinases Targets of the Cyclin–CDK Complexes Triggers for the G1/S and G2/M Transitions

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Protein Degradation Helps Regulate the Cell Cycle 16.4 Checkpoints in the Cell Cycle The DNA Damage Checkpoint The Centrosome Duplication Checkpoint The Spindle Assembly Checkpoint Consequences of Checkpoint Failure 16.5 Cancer Cells Oncogenes and Proto-oncogenes Tumor-Suppressor Genes 16.6 Hereditary Cancer Syndromes Defects in DNA Repair 16.7 Genetics of the Acute Leukemias 16.8 The Genomics and Transcriptomics of Cancer Pancreatic Cancer: The Mutational Landscape Pancreatic Cancer: The Transcriptiosomal Landscape From Genes and Transcripts to Therapy? ROOTS OF DISCOVERY: Cycle-Ops ROOTS OF DISCOVERY: Two Hits, Two Errors

Unit V Variation

17 Mitochondrial DNA and Extranuclear Inheritance Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Learning Objectives & Science Competencies 17.1 Origin and Molecular Genetics of Organelles Organelle Genomes RNA Editing The Genetic Codes of Organelles 17.2 Patterns of Extranuclear Inheritance Maternal Inheritance of Animal Mitochondria Maternal Inheritance and Maternal Effects Heteroplasmy Mitochondrial Genetic Diseases

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Tracing Population History Through Mitochondrial DNA Cytoplasmic Male Sterility in Plants 17.3 Vegetative Segregation Leaf Variegation in Four-O'Clock Plants Respiration-Defective Mitochondrial Mutants in Yeast 17.4 Cytoplasmic Transmission of Symbionts Bacterial Symbionts of Aphids Killer Strains of Paramecium Wolbachia in Arthropods 17.5 Maternal Effect in Snail Shell Coiling ROOTS OF DISCOVERY: A Coming Together ROOTS OF DISCOVERY: How the Aphids Got PVT

18 Genes in Populations Learning Objectives & Science Competencies

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

18.1 Population Genetics Allele Frequencies and Genotype Frequencies Random Mating and the Hardy–Weinberg Principle Implications of the Hardy–Weinberg Principle A Test for Random Mating Frequency of Heterozygous Genotypes Multiple Alleles DNA Typing X-Linked Genes Genetic Variation and Linkage 18.2 Inbreeding The Inbreeding Coefficient Allelic Identity by Descent Calculation of the Inbreeding Coefficient from Pedigrees Effects of Inbreeding

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

18.3 Genetics and Evolution 18.4 Mutation and Migration Irreversible Mutation Reversible Mutation 18.5 Random Genetic Drift Loss of Genetic Variation in Endangered Species 18.6 Natural Selection Selection in a Laboratory Experiment Selection in Diploid Organisms Components of Fitness Selection–Mutation Balance Heterozygote Superiority Molecular Signals of Selection ROOTS OF DISCOVERY: A Yule Message from Dr. Hardy THE CUTTING EDGE: CRISPR-Cas9 for Disease Control?

19 Molecular and Human Evolutionary Genetics Learning Objectives & Science Competencies

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

19.1 Molecular Evolutionary Analysis Gene Trees Bootstrapping Gene Trees and Species Trees Molecular Clock of Evolutionary Change Rates of DNA Evolution Rates of Evolution in Protein-Coding Regions Origins of New Genes: Orthologs and Paralogs 19.2 Ancient DNA 19.3 Where Humans Fit on the Tree of Life Evidence That Humans Are Most Closely Related to Chimpanzees Similarities in Genomic DNA

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Analysis of Multiple Genetic Elements Differences Between Human and Chimpanzee Genomes 19.4 What Do the Genetic Differences Between Humans and Chimpanzees Mean? Molecular Adaptations Unique to Humans FOXP2: A Gene Related to Language Gene-Expression Differences Between Humans and Chimpanzees 19.5 A Synopsis of Human Evolution The Cast of Characters in Human Evolution Models of Modern Human Origins 19.6 Genetic Evidence for Modern Human Origins Tracing Human History Through Mitochondrial DNA The Neandertal Genome The Denisovan Genome 19.7 Measuring Human Diversity

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Tracing Human History with Genetic Markers The Apportionment of Within-Group and BetweenGroup Variation Tracing Human History Through the Y Chromosome 19.8 Genetic Adaptations Unique to Humans Amylase and Dietary Starch Adaptation to Parasites and Disease Evolutionary Adaptation Affecting Human Skin Color THE CUTTING EDGE: The Peopling of Western Europe ROOTS OF DISCOVERY: Starch Contrast

Appendix A: Answers to Even-Numbered Problems Word Roots: Prefixes, Suffixes, and Combining Forms Concise Dictionary of Genetics and Genomics

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Index

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:11.

Preface Students' curiosity about genetics is refreshed by almost daily reports in the media of new discoveries related to genetic differences in drug responses, genetic risk factors for disease, and genetic evidence for human origins and history. They are also intrigued by ethical controversies related to genetics: Should genetic manipulation be used on patients for the treatment of disease? Should human fetuses be used in research? Should human beings be cloned? What are the biological and ethical implications of gene editing? And there are social controversies: Should there be laws governing genetic privacy? Who should have access to genetic testing records, and for what purpose? What is the proper role of direct-to-consumer genotyping services? The teacher of genetics, in turn, faces myriad challenges: ■ Sustaining students' enthusiasm about this topic

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Motivating their desire to understand the principles of genetics in a comprehensive and rigorous way ■ Guiding students in gaining an understanding that genetics is not only a set of principles but also an experimental approach to solving a wide range of biological problems ■ Helping students learn to think about genetic problems and about the wider social and ethical issues arising from genetics and genomics Genetics: Analysis of Genes and Genomes, Ninth Edition, addresses these challenges while emphasizing the beauty, logical

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:22.

clarity, and unity of the subject. Our pedagogical approach is to treat transmission genetics, molecular genetics, and evolutionary genetics as fully integrated subjects. This integration appeals to most modern geneticists, who recognize that the distinctions between the subfields are artificial. The chapters in this text have been arranged into five sections: an overall introduction to genes and genetic analysis, transmission genetics, the organization and replication of DNA, gene expression and regulation, and evolutionary genetics. Recognizing that these topics can be organized in many ways, we have sought to make it possible for individual instructors to customize the order in which they are presented.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Our aim is to provide a clear, comprehensive, rigorous, and balanced introduction to genetics and genomics at the college level. We believe that a good course should maintain the right balance between two important aspects of genetics: (1) genetics as a body of knowledge pertaining to genetic transmission, function, and mutation; and (2) genetics as an experimental approach, or a kit of “tools,” for the study of biological processes such as development or behavior. Any student claiming a knowledge of genetics must achieve the following milestones: ■ Understand the basic processes of gene transmission, mutation, expression, and regulation ■ Be familiar with the principal experimental methods that geneticists and molecular biologists use in their studies, and recognize the advantages and limitations of these approaches ■ Think like a geneticist at the elementary level of being able to formulate genetic hypotheses, work out their consequences, and test the results against observed data

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:22.

■ State genetic principles in his or her own words and recognize the key terms of genetics in context ■ Solve problems of several types, including single-concept exercises that require application of definitions or the basic principles of genetics, problems in genetic analysis in which several concepts must be applied in logical order, and problems in quantitative analysis that call for some numerical calculation ■ Gain some sense of the social and historical context in which genetics and genomics has developed and is continuing to develop

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Acquire a basic familiarity with the genetic resources and information that are available through the Internet

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:08:22.

The Student Experience We have included many special features to help students achieve these learning goals. The text is clearly and concisely written in a student-friendly and relaxed prose style.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Each chapter begins with an explicit statement of the Learning Objectives and Science Competencies that students should aim to achieve. These guideposts are intended to help students identify key concepts and use them at a variety of learning levels, including comprehension, application, analysis, and synthesis. They also serve as powerful study tools when reviewing the material for course and exam preparation.

■ A unique feature of this text is the Roots of Discovery boxes. Each chapter contains one or two of these boxes, in which the text material is connected to excerpts from classic papers that report key experiments in genetics or raise important social, ethical, or legal issues in genetics. Each Roots of Discovery feature includes a brief introduction explaining the importance of the experiment and the historical context in which it was

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

carried out, followed by short excerpts from the original literature of genetics, interspersed with commentary connecting the results and conclusions of the paper to the topics covered in the chapter. Many of these boxes offer excerpts from historical papers, such as Mendel's paper, but

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

by no means are they all “old” papers. Indeed, many of these “classic” papers are very recent.

Some of the pieces were published originally in French, others in German. These excerpts appear in English translation. In papers that use outmoded or unfamiliar terminology, or that use archaic gene symbols, we have substituted the modern equivalent because the use of a consistent terminology in the text and in the Roots of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Discovery features makes the material more accessible to the student. ■ In addition, several chapters contain Cutting Edge boxes, which are designed to introduce students to some of the more innovative work being done in the field. Topics such as the bacterial genetics of CRISPR–Cas9, genome-wide association studies based on personal genomics, and the potential of gene drive are designed to help students appreciate the advances

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

being made at the forefront of genetics. While the Roots of Discovery boxes are designed to highlight the experimental foundations of genetics, the Cutting Edge boxes are intended to give students a taste of where it is going. Each of these features focuses on recent work that has the potential to be transformative in the future or that has introduced new methods of analysis that can be applied to questions in genetics and genomics. Each also includes references to the original work that can be incorporated into class activities and assignments. We hope that the Cutting Edge feature will prove both useful and stimulating, and that the format we have used can be applied to other new advances as they occur.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The art program is spectacular and a learning aid in itself. Every chapter is richly illustrated with beautiful graphics in which color is used functionally to enhance the value of each illustration as a learning aid. The illustrations are also heavily annotated with explanatory boxes explaining in a step-by-step fashion what is happening at each level of the illustration.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

These labels make the art inviting as well as informative. They also allow the illustrations to stand relatively independently of the text, enabling students to review material without rereading the whole chapter. The art program is used not only for its visual appeal but also to increase the pedagogical value of the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

text.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Characteristic colors and shapes have been used consistently throughout the text to indicate different types of molecules—DNA, mRNA, tRNA, and so forth. For example, DNA is illustrated in any one of a number of ways, depending on the level of resolution necessary for the illustration, and each time a particular level of resolution is depicted, the DNA is shown in the same way.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

A great deal of potential confusion is avoided by ensuring that DNA, RNA, and proteins are represented in the same manner in every chapter. Numerous full-color photographs feature molecular models in three dimensions; these give a strong visual reinforcement of the concept of macromolecules as physical entities with defined three-dimensional shapes and charge distributions that serve as the basis of interaction with other macromolecules.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The page design is clean, crisp, and uncluttered. As a result, the text is pleasant to look at and easy to read.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ To help students focus on the key details of this challenging subject, each numbered section concludes with a Summing Up feature that highlights the most important concepts. Then,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

at the end of each chapter, a Chapter Summary in bulleted list form collects key learning points for students. ■ Each chapter concludes with approximately 50 problems for solution, graded in difficulty, for the students to test their understanding. The problems are of several different types.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Review the Basics questions review the major concepts by presenting questions for discussion. These questions ask for genetic principles to be restated in the student's own words; some are matters of definition or call for the application of elementary principles.

The Guide to Problem Solving demonstrates typical problems that apply the principles. The problems are worked in full, showing the concepts needed to solve the problem and the reasoning behind the answer. This feature serves as a review of the important concepts used in working problems. It also highlights some of the most common mistakes made by beginning students and gives pointers on how the student can avoid falling into these conceptual traps.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

A large number of Analysis and Applications problems are provided (with answers to even-numbered problems at the end of the text). In these more traditional types of genetic problems, several concepts must be applied in logical order and some numerical calculation may be required. The level of mathematics is that of arithmetic and elementary probability as it pertains to genetics. None of the problems uses mathematics beyond elementary algebra.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Challenge Problems are similar to the Analysis and Applications problems but are more challenging, often because they require a more extensive analysis of data before the question can be answered.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ At the end of the text are Answers to Even-Numbered Problems, in which the logic of each answer is explained in full. The answers are complete, explaining the logical foundation of the solution and laying out the methods used. We find that many of our students, like students everywhere, often sneak a look at the answer before attempting to solve a problem. This is a pity. Working backward from the answer should be a last resort, because problems are valuable opportunities to learn. Problems that the student cannot solve are usually more important than the ones that the student can solve, because the sticklers usually identify trouble spots, areas of confusion, or gaps in understanding. Therefore, we urge our students to try to answer each question before looking at the answer. The answers to the odd-numbered questions are provided to the instructor as part of the Teaching Tools package so they could be assigned for a grade.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ At the end of the text, we include a short list of Word Roots that are used in many of the technical terms of genetics. The Word Roots are intended to help students interpret and remember the meaning of key terms, and master the essential terminology of genetics and genomics.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The Concise Dictionary of Genetics found at the end of the text enables students to check their understanding of the key words or look up any technical terms they may have forgotten. It includes not only the key words but also genetic terms that students are likely to encounter in exploring the Internet or in their further reading.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved. Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:41.

What's New in the Ninth Edition? This edition has been completely revised and updated. Each

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

chapter has been thoroughly reworked, and the organization of topics, as reflected in the order of chapters, has been revised. To a great extent, this reorganization was driven by developments in the field—especially the explosion of data being generated by highthroughput DNA sequencing, particularly in humans. These advances have revolutionized the study of genetic variation in populations and brought genome-wide association studies (GWAS) and related methods to the forefront. The subject of GWAS is introduced in the context of gene mapping (Chapter 5) and expanded upon with respect to the genetics of complex traits (now Chapter 7). Other applications of high-throughput approaches to genetics include greatly expanded coverage of RNA-seq (Chapter 14), cancer genomics (Chapter 16), and ancient DNA analysis (Chapter 19). Roots of Discovery boxes (formerly Connections boxes) have been rewritten, in most cases reducing the extent of direct quotes from the original papers and providing readers with focused connections to the major points of the topic being covered. All Roots of Discovery boxes contain complete citations of the papers profiled, including URLs, so that they can easily be explored in more depth, either by the interested student or as part of an instructor-designed activity. Cutting Edge boxes, a new feature, also provide connections to the literature to facilitate further exploration.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

Chapter Organization To help students keep track of the main issues and avoid being distracted by details, each chapter opens with an Outline that shows the road map of the territory ahead. This is followed by a list of Learning Objectives and Science Competencies identifying the most important concepts and principles. An opening paragraph gives an overview of the chapter, illustrates the subject with some specific examples, and shows how the material is connected to genetics as a whole. The text makes liberal use of numbered and bulleted lists to help students organize their learning, as well as summary statements set off in special type to emphasize important principles. Each numbered section concludes with a bulleted list entitled “Summing Up,” which highlights the major points covered in the section and will help the student review and master the material covered. Each chapter ends with a Chapter Summary, questions for discussion called Review the Basics, a Guide to Problem Solving, problems for solution in a section called Analysis and Applications, and Challenge Problems.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Contents and Organization Today, most students learn about DNA in grade school or high school; in our teaching, we have found it artificial to pretend that DNA does not exist until the middle of the term. Thus, an important feature is the presence of two introductory chapters providing a broad overview of DNA, genes, and genomes—what they are, how they function, how they change by mutation, and how they evolve through time. The introductory chapters serve to connect the more advanced concepts that students are about to learn with what they already know. They also serve to provide each student with a solid framework for integrating the material that comes later. An important role they play is to provide a sufficient grounding in the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

molecular basis of genetics—DNA structure and replication, as well as the central dogma of molecular biology—so that molecular concepts can be integrated into chapters covering “classical” topics such as Mendelian genetics and gene mapping. Throughout each chapter, there is a balance between observation and theory, between principle and concrete example, and between challenge and motivation. Molecular, classical, and evolutionary genetics are integrated throughout the text. Frequent references are made to human genetics. The text is also liberally populated with applications to other animals and plants, including the key model organisms used in genetics and genomics.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Next, we describe some of the highlights of the five major sections of the text.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

Defining and Working with Genes ■ Genes, Genomes, and Genetic Analysis (Chapter 1) is an overview of genetics designed to bring students with disparate backgrounds to a common level of understanding. This chapter enables classical, molecular, and evolutionary genetics to be integrated in the rest of the text. Included in this chapter are the basic concepts of molecular genetics: DNA structure,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

replication, expression, and mutation. A feature of this chapter is a detailed exploration of the work of Beadle and Tatum that led to the one-gene-one-enzyme hypothesis and the use of complementation analysis as a tool for genetic inference ■ DNA Structure and Genetic Variation (Chapter 2) emphasizes that the primary tools of the modern geneticist are methods for the experimental manipulation of DNA. It includes a more detailed look at DNA structure, and it introduces the principal methods of DNA manipulation, including restriction enzymes, electrophoresis, DNA hybridization, Southern blotting, and the polymerase chain reaction (PCR). We also introduce singlenucleotide polymorphisms (SNPs) and copy-number polymorphisms (CNPs) and discuss how these types of genetic markers can be assayed on a genome-wide scale using oligonucleotide microarrays (DNA chips). The use of simplesequence repeat variation for genotyping is given expanded coverage.

Transmission Genetics ■ Mendelian Genetics: The Principle of Segregation and Assortment (Chapter 3) introduces the fundamental Mendelian concepts of random segregation and independent assortment, not only as they can be inferred from crosses involving experimental organisms like peas and flies, but also by analysis of human pedigrees. The role of probability in genetic

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

analysis is also introduced. A unique feature of this chapter is the integration of molecular genetics with Mendel's experiments. We describe the molecular basis of the wrinkled mutation and show how a modern geneticist would carry out Mendel's study, examining the molecular phenotypes on the one hand and the morphological phenotypes on the other hand. This pedagogy provides a solid basis for understanding not only Mendel's experiments as he actually performed and interpreted them, but also the use of modern molecular approaches in genetic analysis. Molecular markers are also integrated into human genetic analysis.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The Chromosomal Basis of Inheritance (Chapter 4) covers the cell cycle, mitosis and meiosis, and hypothesis testing in genetics. The approach to the chi-squared test has been revised by introducing random variables and their distributions, first with respect to the binomial distribution and then with respect to the chi-squared distribution. Computer-simulated distributions are used to illustrate the logic of significance testing, after which the mechanics of actually performing a chisquared test are described. ■ Genetic Linkage and Chromosome Mapping (Chapter 5) covers linkage analysis and gene mapping. The classical approach, involving test crosses in experimental organisms, has been retained, as has streamlined coverage of tetrad analysis in fungi. New to the ninth edition is a section on genetic mapping in humans, which includes expanded coverage of lod score analysis of pedigrees, as well as an entirely new section on gene localization by association analysis. ■ Human Karyotypes and Chromosome Behavior (Chapter 6) covers the principles of chromosome mechanics with special reference to human chromosome number and structure and the types of aberrations that are found in human chromosomes.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

The genetic implications of chromosome abnormalities— duplications, deficiencies, inversions, and translocations—are also discussed. The evolutionary significance of gene duplication is introduced, and the genetic consequences of translocation heterozygosity (pseudolinkage) is described. ■ The Genetic Basis of Complex Traits (Chapter 7) covers quantitative genetics. It is a revision of Chapter 18 from the previous edition. We chose to introduce this material earlier

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

because of the growing importance of genome-wide association studies in both agriculture and medicine. Crohn's disease is used as a model for doing so, and both the strengths and weaknesses of GWAS as a tool are highlighted. Coverage of heritability has been revised, and a new section on misconceptions about it has been added. ■ Genetics of Bacteria and Their Viruses (Chapter 8) extends the use of genetic analysis to bacteria (mainly Escherichia coli) and viruses, and serves as a transition into the more molecular subjects covered in subsequent chapters. The section on plasmids has been revised to focus more specifically on the F plasmid; more detailed coverage of bacterial plasmids in general has been moved to the Molecular Organization of Chromosomes and Genomes chapter. Some of the classical experiments and methods, such as replica plating and the “U-tube” experiment, are highlighted. The description of the origin of the term “cistron” and the cis-trans test has been expanded, and CRISPR–Cas9 as a bacterial system is introduced in a Cutting Edge box.

Organization and Replication of Chromosomes and DNA ■ Molecular Organization of Chromosomes and Genomes (Chapter 9) continues the transition into the molecular basis of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

inheritance, covering the organization of DNA in the genomes of both prokaryotes and eukaryotes. It now includes an introduction to mobile DNA in both types of organisms. The section on Cot kinetics has been shortened, recognizing that while historically and conceptually important, this methodology is no longer in active use. The description of chromosome condensation has been updated, and its connection to mitosis (and meiosis) has been made more explicit.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ DNA Replication and Sequencing (Chapter 10) covers the mechanics of DNA replication and methods of sequencing as the starting point for an in-depth discussion of the molecular basis of genetics. The mechanisms of DNA replication are described first for prokaryotes and then for eukaryotes. Coverage of Sanger sequencing is followed by a description of sequencing-by-synthesis, using the Illumina platform as a model. ■ Mutation, Repair, and Recombination (Chapter 11) covers mutation, DNA repair, and recombination. In previous editions, mutation was presented in a later chapter, but by having it immediately follow coverage of DNA replication, we hope to emphasize the importance of all of these processes in both the maintenance of genetic continuity and the generation of new variation. Coverage of the classical Luria–Delbruck experiment, showing that mutation is random with respect to adaptation, has been added. Coverage of mechanisms of recombination has been incorporated in this chapter, emphasizing the similarity between these mechanisms and post-replication repair mechanisms.

Gene Expression ■ Molecular Biology of Gene Expression (Chapter 12) covers the processes of transcription, translation, and RNA

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

processing. Transcription mechanisms in prokaryotes and eukaryotes are more clearly differentiated; the concept of consensus sequences is introduced more comprehensively, along with visualizations of them as “sequence logos.” ■ Molecular Mechanisms of Gene Regulation (Chapter 13) covers the regulation of gene expression. Mechanisms in prokaryotes and eukaryotes have been separated, so that they can be better compared and contrasted. The section on the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

role played by noncoding RNAs in regulation has been expanded, highlighting their roles in processes such as Xchromosome inactivation, alternative mRNA splicing, and translational regulation. ■ Manipulating Genes and Genomes (Chapter 14) covers experimental methods of gene manipulation, beginning with the classical methods of gene cloning and proceeding forward to genomic analysis, transcriptomics, and gene editing with CRISPR–Cas9. Included are the use of restriction enzymes and vectors in recombinant DNA, cloning strategies, sitedirected mutagenesis, the production of genetically defined transgenic animals and plants, and applications of genetic engineering. Coverage of whole-exome sequencing has been expanded, and a new section on functional genomics has been added. This material includes expanded coverage of RNA-seq and a description of quantitative PCR for transcript quantification. A subsection on use of RNAi for gene expression knockdown has been added, as has a major section on gene editing with CRISPR–Cas9. Associated with the latter is a Cutting Edge box on the use of these methods in a mouse model of muscular dystrophy. ■ Genetic Control of Development (Chapter 15) focuses on genetic analysis of development in nematodes (Caenorhabditis elegans) and Drosophila and includes a thorough examination

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

of the genetic basis of floral development in Arabidopsis thaliana. A new section on the role of regulatory RNAs in the control of development has been added, including examples from nematodes, flies, and mammalian Hox genes. ■ Molecular Genetics of the Cell Cycle and Cancer (Chapter 16) investigates cancer from the standpoint of the genetic control of cell division, with emphasis on the checkpoints that, in normal cells, result either in inhibition of cell division or in programmed cell death (apoptosis). Cancer results from a series of successive mutations, usually in somatic cells that overcome the normal checkpoints that control cellular proliferation. A new section on cancer genomics has been added, using pancreatic cancer as a model for characterization of the mutational and transcriptisomal changes that are associated with tumor progression.

Variation

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Mitochondrial DNA and Extranuclear Inheritance (Chapter 17) covers non-nuclear genetics, including genetic defects in human mitochondrial DNA. A new section has been added that describes the transmission of Wolbachia in arthropods. ■ Genes in Populations (Chapter 18) covers population genetics (material on molecular evolution has been moved to the Molecular and Human Evolutionary Genetics chapter). The section on DNA fingerprinting has been updated to illustrate use of short tandem repeats as markers. A new subsection on linkage disequilibrium has been added, along with “heat map” illustrations of disequilibrium derived from genomic data. In the discussion of evolutionary processes, the order of topics has been rearranged so that genetic drift—a random process found in all finite populations—is addressed prior to examining the deterministic process of natural selection. A new section on

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

molecular signals of selection has been added, incorporating material on lactase persistence in humans. A Cutting Edge box describes the principles of “gene drive” based on CRISPR– Cas9 and explains how it might be used to control mosquito vectors of malaria. ■ Molecular and Human Evolutionary Genetics (Chapter 19) covers both molecular and human evolution, since most of the advances in the latter have resulted from developments in the former. Other changes in this chapter include an expanded section on ancient DNA analysis and the use of Alu1 insertion sites to resolve hominid phylogeny. Coverage of diversity in the genus Homo has been revised to include new findings about H. denisova and H. floresiensis. The section on genetic variation in H. sapiens has been revised to incorporate data from the 1000 Genomes Project (as opposed to HapMap in previous editions) and to include the use of genetic assignment methods (“STRUCTURE”) for characterization of modern human variation. A new Cutting Edge box focuses on use of ancient DNA genotyping to characterize the peopling of Western Europe.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Flexibility There is no requirement to start at the beginning of this text and proceed straight to the end. In fact, doing so in a typical onesemester course borders on the impossible. In this ninth edition, each chapter has been designed to be a self-contained unit that stands on its own. This feature gives instructors the option of using whatever order suits them, and of skipping particular chapters that do not fit with their course design. We have integrated molecular and classical principles throughout the text, so you can begin a course with almost any of the chapters. Most instructors will prefer to start with the overview in the Genes, Genomes, and Genetic

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

Analysis chapter because it brings every student to the same basic level of understanding. The DNA Structure and Genetic Variation chapter introduces the basic experimental manipulations used in modern genetics and serves to integrate molecular and classical genetics in the discussion of Mendel in the Transmission Genetics: The Principle of Segregation chapter. Some other approaches that instructors might use to structure the beginning of their course are outlined here.

Approach

Chapters to begin with

Mendel-early format

Genes, Genomes, and Genetic Analysis Transmission Genetics: The Principle of Segregation DNA Structure and Genetic Variation

Chromosome-early format

Genes, Genomes, and Genetic Analysis Chromosomes and Sex-Chromosome Inheritance DNA Structure and Genetic Variation Transmission Genetics: The Principle of Segregation

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Genetic Linkage and Chromosome Mapping

Genomes-first format

Genes, Genomes, and Genetic Analysis DNA Structure and Genetic Variation Chromosomes and Sex-Chromosome Inheritance

The writing and illustration program were designed to accommodate a variety of formats, and we encourage teachers to take advantage of this flexibility to meet their own unique needs.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved. Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:09:52.

Teaching Tools Jones & Bartlett Learning offers a suite of traditional and interactive multimedia supplements to assist instructors and aid students in mastering genetics. Additional information and review copies of any of the following items are available through your Jones & Bartlett Learning sales representative.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The PowerPoint Image Bank is an easy-to-use multimedia tool that provides all of the illustrations and photos from the text (to which Jones & Bartlett Learning holds the rights to reproduce electronically) for use in classroom presentation. You can select images you need or easily generate your own slide shows, or you can print the files for transparency creation. Many images have already been inserted into the PowerPoint Lecture Outline presentations for ease of use.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:11.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ A PowerPoint presentation containing the detailed outline for each chapter of Genetics: Analysis of Genes and Genomes, Ninth Edition, is included. This presentation, which is designed

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:11.

to mirror the text, is constructed flexibly to meet your lecture's

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

organization. The outline is open, allowing you to provide the elements you deem necessary, whether it be new text or more images from the Image Bank.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:11.

■ A Test Bank containing more than 1,000 questions and complete answers is available. The questions are a mix of factual, descriptive, and quantitative types. A typical chapter contains multiple choice, fill-in-the-blank, and short-answer questions. The Test Bank is provided as Microsoft Word documents.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The Instructor's Manual includes Chapter Summaries, Teaching Tips, References, Suggested Activities, and more.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:11.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved. Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:11.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved. Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:11.

Acknowledgments We are indebted to many colleagues whose advice, thoughtful reviews, and comments have helped in the preparation of this and previous editions. Their expert recommendations are reflected in the content, organization, and presentation of the material. Laura Adamkewicz, George Mason University, Fairfax, VA Jeremy C. Ahouse, Brandeis University, Waltham, MA Mary Alleman, Duquesne University, Pittsburgh, PA Peter D. Ayling, University of Hull, Hull, United Kingdom John C. Bauer, Stratagene, Inc., La Jolla, CA Anna W. Berkovitz, Purdue University, West Lafayette, IN Mary K. B. Berlyn, Yale University, New Haven, CT Justin Blumenstiel, University of Kansas, Lawrence, KS Thomas A. Bobik, University of Florida, Gainesville, FL David Botstein, Princeton University, Princeton, NJ

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Colin G. Brooks, The Medical School, Newcastle, United Kingdom Pierre Carol, Université Joseph Fourier, Grenoble, France Domenico Carputo, University of Naples, Naples, Italy Sean Carroll, University of Wisconsin, Madison, WI Chris Caton, University of Birmingham, Birmingham, AL John Celenza, Boston University, Boston, MA Estelle Chamoux, Bishop's University, Sherbrooke, Quebec, Canada Alan C. Christensen, University of Nebraska–Lincoln, Lincoln, NE

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:22.

Christoph Cremer, University of Heidelberg, Heidelberg, Germany Marion Cremer, Ludwig Maximilians University, Munich, Germany Thomas Cremer, Ludwig Maximilians University, Munich, Germany Leslie Dendy, University of New Mexico, Los Alamos, NM Stephen J. DíSurney, University of Mississippi, Oxford, MS John W. Drake, National Institute of Environmental Health Sciences, Research Triangle Park, NC Kathleen Dunn, Boston College, Boston, MA Chris Easton, State University of New York, Binghamton, NY Joel C. Eissenberg, St. Louis University, St. Louis, MO Ioannis Eleftherianos, The George Washington University, Washington, DC Wolfgang Epstein, University of Chicago, Chicago, IL Brian E. Fee, Manhattan College, Riverdale, NY Gyula Ficsor, Western Michigan University, Kalamazoo, MI Robert G. Fowler, San Jose State University, San Jose, CA David W. Francis, University of Delaware, Newark, DE

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Matthew K. Fujita, University of Texas at Arlington, Arlington, TX James Fuller, University of North Carolina, Chapel Hill, NC Gail Gasparich, Towson University, Towson, MD Elliott S. Goldstein, Arizona State University, Tempe, AZ Kent G. Golic, University of Utah, Salt Lake City, UT Lesley H. Greene, Old Dominion University, Norfolk, VA Patrick Guilfoile, Bemidji State University, Bemidji, MN Jeffrey C. Hall, Brandeis University, Waltham, MA Mark L. Hammond, Campbell University, Buies Creek, NC

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:22.

Steven Henikoff, Fred Hutchinson Cancer Research Center, Seattle, WA Charles Hoffman, Boston College, Boston, MA Ivan Huber, Fairleigh Dickinson University, Madison, NJ Kerry Hull, Bishop's University, Quebec, Canada Lynn A. Hunter, University of Pittsburgh, Pittsburgh, PA Richard Imberski, University of Maryland, College Park, MD Bradley Jacob Isler, Ferris State University, Big Rapids, MI Nitya P. Jacob, Oxford College of Emory University, Oxford, GA Joyce Katich, Monsanto, Inc., St. Louis, MO Jeane M. Kennedy, Monsanto, Inc., St. Louis, MO Jeffrey King, University of Bern, Bern, Switzerland Anita Klein, University of New Hampshire, Durham, NH Tobias A. Knoch, German Cancer Research Center, Heidelberg, Germany Yan B. Linhart, University of Colorado, Boulder, CO K. Brooks Low, Yale University, New Haven, CT Elena R. Lozovsky, Harvard University, Cambridge, MA Sally A. MacKenzie, Purdue University, West Lafayette, IN Irina Makarevitch, Hamline University, Saint Paul, MN Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Gustavo Maroni, University of North Carolina, Chapel Hill, NC Lauren McIntyre, University of Florida, Gainesville, FL Eric Mendenhall, University of Alabama in Huntsville, Huntsville, AL Jeffrey Mitton, University of Colorado, Boulder, CO Robert K. Mortimer, University of California, Berkeley, CA Gisela Mosig, Vanderbilt University, Nashville, TN Erin W. Norcross, Mississippi College, Clinton, MS Steve O'Brien, National Cancer Institute, Frederick, MD

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:22.

Daniel Odom, California State University, Northridge, Northridge, CA Kevin O'Hare, Imperial College, London, United Kingdom Ronald L. Phillips, University of Minnesota, St. Paul, MN David Pilbeam, Harvard University, Cambridge, MA Robert Pruitt, Purdue University, West Lafayette, IN Pamela Reinagel, California Institute of Technology, Pasadena, CA Rebecca Reiss, New Mexico Tech, Socorro, NM Susanne Renner, University of Missouri, St. Louis, MO Michael Robinson, Miami University, Oxford, OH Andrew J. Roger, Dalhousie University, Halifax, Nova Scotia, Canada Kenneth E. Rudd, National Library of Medicine, Bethesda, MD Thomas F. Savage, Oregon State University, Corvallis, OR Joseph Schlammandinger, University of Debrecen, Hungary Julia M. Schmitz, Piedmont College, Athens, GA Barbara B. Sears, Michigan State University, East Lansing, MI David Shepard, University of Delaware, Newark, DE Alastair G. B. Simpson, Dalhousie University, Halifax, Nova Scotia, Canada

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Navin K. Sinha, Rutgers University, Piscataway, NJ Leslie Smith, National Institute of Environmental Health Sciences, Research Triangle Park, NC Charles Staben, University of Kentucky, Lexington, KY Mark Sturtevant, Oakland University, Rochester, MI Johan H. Stuy, Florida State University, Tallahassee, FL David T. Sullivan, Syracuse University, Syracuse, NY Jeanne Sullivan, West Virginia Wesleyan College, Buckhannon, WV

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:22.

Millard Susman, University of Wisconsin, Madison, WI Barbara Taylor, Oregon State University, Corvallis, OR Irwin Tessman, Purdue University, West Lafayette, IN Asli Tolun, Bogazici University, Istanbul, Turkey Yoshi Tomoyasu, Miami University, Oxford, OH Maria Tsompana, University of North Carolina, Chapel Hill, NC Micheal Tully, University of Bath, Bath, United Kingdom David Ussery, The Technical University of Denmark, Lyngby, Denmark George von Dassow, Friday Harbor Laboratories, Friday Harbor, WA Denise Wallack, Muhlenberg College, Allentown, PA Kenneth E. Weber, University of Southern Maine, Gorham, ME Tamara Western, Okanagan University College, Kelowna, British Columbia, Canada Darla J. Wise, Concord University, Athens, WV Jay Zimmer, Gardner-Webb University, Boiling Springs, NC

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

We also wish to acknowledge Barbara B. Sears of Michigan State University for her many helpful suggestions for the chapter on organelle genetics in previous editions of this text. We also wish to acknowledge the superb editorial and production staff of Jones & Bartlett Learning who helped make this text possible: Matt Kane, Audrey Schwinn, Loren-Marie Durr, Nancy Hitchcock, Wes DeShano, Troy Liston, and Scott Moden. Much of the credit for the attractiveness and readability of the text should go to them. In addition, we thank Jones & Bartlett Learning for its ongoing commitment to high-quality work in book production. We are grateful to the many people, acknowledged in the legends of the figures, who contributed photographs, drawings, and micrographs from their own research and publications.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:22.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved. Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:22.

About the Authors Daniel L. Hartl is Higgins Professor of Biology at Harvard University and a member of the National Academy of Sciences and the American Academy of Arts and Sciences. He received his BS and PhD from the University of Wisconsin and carried out postdoctoral research at the University of California at Berkeley. His research interests include molecular genetics, genomics, molecular evolution, and population genetics.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Bruce J. Cochrane is Professor of Biology at Miami University. He received his BA in biology at Cornell University and his PhD in genetics at Indiana University, and carried out postdoctoral research at the University of North Carolina. His research interests include population genetics, evolutionary biology, and bioinformatics.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:32.

© Molekuul/Science Photo Library/Getty Images.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

UNIT 1 Defining and Working with Genes Chapter 1 Genes, Genomes, and Genetic Analysis Chapter 2 DNA Structure and Genetic Variation

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:10:32.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Medical Images RM/Jane Hurd.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

CHAPTER 1 Genes, Genomes, and Genetic Analysis CHAPTER OUTLINE 1.1 DNA as the Genetic Material 1.2 DNA Structure and Replication 1.3 Genes and Proteins 1.4 Genetic Analysis 1.5 Gene Expression: The Central Dogma 1.6 Mutation and Variation 1.7 Genes and Environment 1.8 The Molecular Unity of Life ROOTS OF DISCOVERY: The Black Urine Disease

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Archibald E. Garrod (1908) Inborn Errors of Metabolism ROOTS OF DISCOVERY: One Gene, One Enzyme George W. Beadle and Edward L. Tatum (1941) Genetic Control of Biochemical Reactions in Neurospora

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

LEARNING OBJECTIVES & SCIENCE COMPETENCIES Once you have understood and can apply the principles of genes, genomes, and genetic analysis discussed in this chapter, you will have acquired the following science competencies: ■ Given the base sequence of a transcribed strand of protein-coding DNA, specify the sequence of bases in the corresponding region of messenger RNA and the corresponding sequence of amino acids in the protein. ■ For a mutation in the DNA in which one base is replaced with another, show how the mRNA and protein will be altered.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Given a linear metabolic pathway for an essential nutrient, determine which intermediates will restore the ability to grow to mutant strains that are defective for any of the enzymes in the pathway. ■ Interpret data specifying which intermediates in a linear metabolic pathway restore the ability of mutants to grow, so as to determine the order in which the enzymes and the intermediates appear in the pathway. ■ Given data on the complementation or lack of complementation among all pairs of a set of mutations affecting a biological process, categorize the mutations into groups, each corresponding to a different gene.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

ach species of living organism has a unique set of inherited

E

characteristics that makes it different from every other species. Each species has its own developmental plan— often described as a sort of “blueprint” for building the organism—which is encoded in the DNA molecules present in

its cells. This developmental plan determines the characteristics that are inherited. Because organisms in the same species share the same developmental plan, organisms that are members of the same species usually resemble one another. For example, it is

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

easy to distinguish a human being from a chimpanzee or a gorilla. A human being habitually stands upright and has long legs, relatively little body hair, a large brain, and a flat face with a prominent nose, jutting chin, distinct lips, and small teeth. All of these traits are inherited—part of our developmental plan—and help set us apart as Homo sapiens. But human beings are by no means identical. Many traits, or observable characteristics, differ from one person to another. In addition to notable differences between males and females, there is a great deal of variation in hair color, eye color, skin color, height, weight, personality traits, and other characteristics. Some human traits (such as sex) are transmitted biologically, others culturally. The color of our eyes results from biological inheritance, but the native language we learned as a child results from cultural inheritance. Many traits are influenced jointly by biological inheritance and environmental factors. For example, weight is determined in part by inheritance but also in part by environment: how much food we eat, its nutritional content, our exercise regimen, and so forth. Genetics is the study of biologically inherited traits, including traits that are influenced in part by the environment. Genomics is the study of all the genes in an organism to understand their molecular

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

organization, function, interaction, and evolutionary history. The fundamental concept of genetics and genomics is as follows: Inherited traits are determined by the elements of heredity that are transmitted from parents to offspring in reproduction; these elements of heredity are called genes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The existence of genes and the rules governing their transmission from generation to generation were first articulated by Gregor Mendel in 1866 (described in the chapter titled Mendelian Genetics: The Principles of Segregation and Assortment). Mendel's formulation of inheritance was stated in terms of the abstract rules by which hereditary elements (he called them “factors”) are transmitted from parents to offspring. His objects of study were garden peas, with variable traits such as pea color and plant height. Mendel thus introduced studying genetics through analysis of the progeny produced from matings. Genetic differences between species were impossible to define, because organisms of different species usually do not mate; if they do mate, they typically produce hybrid progeny that die or are sterile. The approach to the study of genetics through the analysis of the offspring from matings is often referred to as classical genetics, or organismic or morphological genetics. By contrast, molecular genetics is the study of the chemical nature of genes and their products. Advances in this area have made it possible to study gene structure and function, as well as differences between species, through the comparison and analysis of the DNA itself. Nevertheless, there is no fundamental distinction between classical and molecular genetics. Rather, the two are different and complementary ways of studying the same thing: the function of the genetic material. This text includes many examples

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

showing how molecular and classical genetics can be used in combination to enhance the power of genetic analysis. In the rest of this chapter, we will focus on groundbreaking observations and experiments that form the basis for our current understanding of the structure and function of genes. Before we do so, however, we need to consider the four necessary properties of genetic material:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1. The genetic material must encode information 2. That information must be used to direct the functioning of cellular processes. 3. The information contained must be transmissible from cell to cell and from generation to generation. 4. There must be the potential for mutation that can result in the physical variation that exists among individuals and between species.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.1 DNA as the Genetic Material The foundation of genetics as a molecular science dates back to 1869, just three years after Mendel reported his experiments. It was in that year that Friedrich Miescher discovered a new type of weak acid, abundant in the nuclei of white blood cells. Miescher's weak acid turned out to be the chemical substance we now call DNA (deoxyribonucleic acid). For many years, the biological

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

function of DNA was unknown, and no role in heredity was ascribed to it. This section describes how DNA was eventually isolated and identified as the material of which genes are made. That the cell nucleus plays a key role in inheritance was recognized in the 1870s when scientists observed that the nuclei of male and female reproductive cells undergo fusion in the process of fertilization. Soon thereafter, chromosomes were first observed inside the nucleus as thread-like objects that become visible in the light microscope when the cell is stained with certain dyes. Chromosomes were found to exhibit a characteristic “splitting” behavior in which each daughter cell formed by cell division receives an identical complement of chromosomes (see the chapter titled The Chromosomal Basis of Inheritance). Further evidence for the importance of chromosomes was provided by the observation that, whereas the number of chromosomes in each cell may differ among biological species, the number of chromosomes is nearly always constant within the cells of any particular species. These features of chromosomes were well understood by about 1900, and they made it seem likely that chromosomes were the carriers of the genes.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

© Science Source.

By the 1920s, several lines of indirect evidence began to suggest a close relationship between chromosomes and DNA. Microscopic studies with special stains showed that DNA is present in chromosomes. Chromosomes also contain various types of proteins, but the amount and kinds of chromosomal proteins differ greatly from one cell type to another, whereas the amount of DNA per cell is constant. Furthermore, nearly all of the DNA present in cells of higher organisms is present in the chromosomes. These arguments for DNA as the genetic material were unconvincing, however, because crude chemical analyses had suggested (erroneously, as it turned out) that DNA lacks the chemical diversity needed to encode the complex set of instructions necessary to build even a simple cell or organism. The favored candidate for the genetic material was protein, because proteins were known to be an exceedingly diverse collection of molecules. Proteins therefore became widely accepted as the genetic material, and DNA was assumed to function merely as the structural framework of the chromosomes. The experiments described in this section showed that, in fact, DNA is the genetic material.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Experimental Proof of the Genetic Function of DNA In 1928, Frederick Griffith took an important first step in elucidating DNA's role when he demonstrated that genetic material can be transferred from one bacterial cell to another. Griffith was working with two strains of the bacterium Streptococcus pneumoniae identified as S and R. When a bacterial cell is grown on solid

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

medium, it undergoes repeated cell divisions to form a visible clump of cells called a colony. The S type of S. pneumoniae synthesizes a gelatinous capsule composed of complex carbohydrate (polysaccharide). The enveloping capsule makes each colony large and gives it a glistening or smooth (S) appearance (FIGURE 1.1). This capsule also enables the bacterium to cause pneumonia, by protecting it from the defense mechanisms of an infected animal. The R strains of S. pneumoniae are unable to synthesize the capsular polysaccharide; they form small colonies that have a rough (R) surface (Figure 1.1). The R strain of the bacterium does not cause pneumonia, because without the capsule the bacteria are inactivated by the immune system of the host. Both types of bacteria “breed true,” in the sense that the progeny formed by cell division have the capsular type of the parent, either S or R.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

FIGURE 1.1 Colonies of rough (R, the small colonies) and smooth (S, the large colonies) strains of Streptococcus pneumoniae. The S colonies are larger because of the gelatinous capsule on the S cells. Photograph reproduced from Avery, O.T., MacLeod, C.M., & McCarty, M. (1944). Studies on the chemical nature of the substance inducing transformation of pneumococcal types. Journal of Experimental Medicine, 79(2), 137–158. Copyright 1994 by Rockefeller University Press. Reproduced with permission of Rockefeller University Press. doi:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

10.1084/jem.79.2.137.

Mice injected with living S cells get pneumonia. In contrast, mice injected with either living R cells or heat-killed S cells remain healthy. Griffith's critical finding was that mice injected with a mixture of living R cells and heat-killed S cells contract the disease and often die of pneumonia (FIGURE 1.2). Bacteria isolated from blood samples of these dead mice produce S cultures with a capsule typical of the injected S cells, even though the injected S cells had been killed by heat. Evidently, the injected material from the dead S cells includes a substance that can be transferred to living R cells, which then enables the R cells to synthesize the Stype capsule. In other words, the R bacteria can be changed—or undergo transformation—into S bacteria. Furthermore, the ability

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

to synthesize the capsule is inherited by descendants of the transformed bacteria.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.2 Griffith's experiment demonstrating bacterial transformation. A mouse remains healthy if injected with either the nonvirulent R strain of S. pneumoniae or heat-killed cell fragments of the usually virulent S strain. R cells in the presence of heat-killed S cells are transformed into the virulent S strain, causing pneumonia in the mouse.

Transformation in Streptococcus was originally discovered in 1928, but it was not until 1944 that the chemical substance responsible for changing the R cells into S cells was identified. In a milestone experiment, Oswald Avery, Colin MacLeod, and Maclyn McCarty showed that the substance causing the transformation of R cells into S cells was DNA. In doing their experiments, these researchers first needed to develop chemical procedures for isolating almost pure DNA from cells, which had never been done before. When they added DNA isolated from S cells to growing cultures of R cells, they observed transformation—that is, a few S cells were produced. Although the DNA preparations contained

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

traces of protein and RNA (ribonucleic acid, an abundant cellular macromolecule chemically related to DNA), the transforming activity was not altered by treatments that destroyed either protein or RNA. Conversely, treatments that destroyed DNA eliminated the transforming activity (FIGURE 1.3). These experiments implied that

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the substance responsible for genetic transformation was the DNA of the cell—an hence that DNA is the genetic material.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.3 A diagram of the Avery–MacLeod–McCarty experiment demonstrating that DNA is the active material in bacterial transformation. (A) Purified DNA extracted from heatkilled S cells can convert some living R cells into S cells, but the material may still contain undetectable traces of protein and/or RNA. (B) The transforming activity is not destroyed by either protease or RNase. (C) The transforming activity is destroyed by DNase and so probably consists of DNA.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Genetic Role of DNA in Bacteriophages Another pivotal finding was reported by Alfred Hershey and Martha Chase in 1952. They studied cells of the intestinal bacterium Escherichia coli after infection by the virus T2. A virus that attacks bacterial cells is called a bacteriophage, a term often shortened to phage. Bacteriophage means “bacteria-eater.” The structure of a

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

bacteriophage T2 particle is illustrated in FIGURE 1.4. This particle is exceedingly small, yet it has a complex structure composed of head (which contains the phage DNA), collar, tail, and tail fibers. (The head of a human sperm is about 30–50 times larger in both length and width than the head of T2.) Hershey and Chase were already aware that T2 infection proceeds via the attachment of the tip of a phage particle's tail to the bacterial cell wall, entry of phage material into the cell, multiplication of this material to form a hundred or more progeny phage, and release of the progeny phage by bursting (lysis) of the bacterial host cell. They also knew that T2 particles were composed of DNA and protein in approximately equal amounts.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

FIGURE 1.4 (A) Drawing of E. coli phage T2, showing various components. The DNA is confined to the interior of the head. (B) An electron micrograph of phage T4, a closely related phage. Electron micrograph courtesy of Robert Duda, University of Pittsburgh.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Because DNA contains phosphorus but no sulfur, whereas most proteins contain sulfur but no phosphorus, it is possible to label DNA and proteins differentially by using radioactive isotopes of the two elements. Hershey and Chase produced particles containing radioactive DNA by infecting E. coli cells that had been grown for several generations in a medium that included 32P (a radioactive isotope of phosphorus) and then collecting the phage progeny. Other particles containing labeled proteins were obtained in the same way, by using a medium that included 35S (a radioactive isotope of sulfur). In the experiments summarized in FIGURE 1.5, nonradioactive E. coli cells were infected with phage labeled with either 32P (part A) or 35S (part B) so that the DNA and proteins could be followed on their individual paths. Infected cells were separated from unattached phage particles by centrifugation, resuspended in fresh

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

medium, and then swirled violently in a kitchen blender to shear attached phage material from the cell surfaces. This treatment was found to have no effect on the subsequent course of the infection, which implies that the phage genetic material must enter the infected cells very soon after phage attachment. The kitchen blender turned out to be the critical piece of equipment. Other methods had been tried to tear the phage heads from the bacterial cell surface, but nothing had worked reliably. Hershey later explained, “We tried various grinding arrangements, with results

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

that weren't very encouraging. When Margaret McDonald loaned us her kitchen blender, the experiment promptly succeeded.”

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.5 The Hershey–Chase (“blender”) experiment demonstrating that DNA, not protein, is responsible for directing the reproduction of phage T2 in infected E. coli cells. (A) Radioactive DNA is transmitted to progeny phage in substantial amounts. (B) Radioactive protein is transmitted to progeny phage in negligible amounts.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

After the phage heads were removed by the blender treatment, the infected bacteria were examined. Most of the radioactivity from 32P-labeled phage was found to be associated with the bacteria, whereas only a small fraction of the 35S radioactivity was present in the infected cells. The retention of most of the labeled DNA, in contrast to the loss of most of the labeled protein, implied that a T2 phage transfers most of its DNA, but very little of its protein, to the cell it infects. The critical finding (Figure 1.5) was that about 50 percent of the transferred 32P-labeled DNA, but less than 1 percent of the transferred 35S-labeled protein, was inherited by the progeny

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

phage particles. Hershey and Chase interpreted this result to mean that the genetic material in T2 phage is DNA. The experiments of Avery, MacLeod, and McCarty and those of Hershey and Chase are regarded as classics in the demonstration that genes consist of DNA. At the present time, the equivalent of the transformation experiment is carried out daily in many research laboratories throughout the world, usually with bacteria, yeast, or animal or plant cells grown in culture. These experiments indicate that DNA is the genetic material in these organisms as well as in phage T2. Although there are no known exceptions to the generalization that DNA is the genetic material in all cellular organisms and many viruses, in several types of viruses the genetic material consists of RNA. These RNA-containing viruses include the human immunodeficiency virus HIV-1 that causes AIDS (acquired immunodeficiency syndrome).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

SUMMING UP ■ DNA was identified as a component of chromosomes, but its significance as the genetic material was initially not recognized. ■ Frederick Griffith showed that genetic information could be transferred in the process now known as transformation. ■ Avery, McLeod, and McCarty demonstrated that the “transforming principle” was DNA.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Hershey and Chase showed that DNA—but not protein—is transferred to host cells during phage infection.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.2 DNA Structure and Replication The inference that DNA is the genetic material left many questions unanswered. How is the DNA in a gene duplicated when a cell divides? How does the DNA in a gene control a hereditary trait? What happens to the DNA when a mutation (a change in the DNA) takes place in a gene?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In the early 1950s, a number of researchers began to try to elucidate the detailed molecular structure of DNA in hopes that the structure alone would suggest answers to these questions. In 1953, James Watson and Francis Crick at Cambridge University proposed the first essentially correct three-dimensional structure of the DNA molecule. The structure was dazzling in its elegance and revolutionary in suggesting how DNA duplicates itself, controls hereditary traits, and undergoes mutation. Even while their tin-andwire model of the DNA molecule was still incomplete, Crick would visit his favorite pub and exclaim “we have discovered the secret of life.” In the Watson–Crick structure, DNA consists of two long chains of subunits, each twisted around the other to form a double-stranded helix. The double helix is right-handed, which means that as one looks along the barrel, each chain follows a clockwise path as it progresses. You can visualize the right-handed coiling in part A of FIGURE 1.6 if you imagine yourself looking up into the structure from the bottom. The dark spheres outline the “backbone” of each individual strand, and they coil in a clockwise direction. The subunits of each strand are nucleotides, each of which contains any one of four chemical constituents called bases attached to a phosphorylated molecule of the five-carbon sugar deoxyribose. The four bases in DNA are:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

■ Adenine (A) ■ Guanine (G) ■ Thymine (T) ■ Cytosine (C)

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.6 Molecular structure of the DNA double helix in the standard “B form.” (A) A space-filling model, in which each atom is depicted as a sphere. (B) A diagram highlighting the helical strands around the outside of the molecule and the A–T and G–C base pairs inside.

The chemical structures of the nucleotides and bases need not concern us at this time. They are examined in the DNA Structure and Genetic Variation chapter. A key point for our present purposes is that the bases in the double helix are paired as shown in Figure 1.6B:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

At any position on the paired strands of a DNA molecule: ■ If one strand has an A, then the partner strand has a T. ■ If one strand has a G, then the partner strand has a C. The pairing between A and T and between G and C is said to be complementary; the complement of A is T, and the complement of G is C. The complementary pairing means that each base along one strand of the DNA is matched with a base in the opposite position on the other strand. Furthermore: Nothing restricts the sequence of bases in a single strand, so any sequence could be present along one strand.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

This principle explains how only four bases in DNA can code for the huge amount of information needed to make an organism. It is the sequence of bases along the DNA that encodes the genetic information, and the sequence is completely unrestricted. The complementary pairing is also called Watson–Crick pairing. In the three-dimensional structure in Figure 1.6A, the base pairs are represented by the lighter spheres filling the interior of the double helix. The base pairs lie almost flat, stacked on top of one another perpendicular to the long axis of the double helix, like pennies in a roll. When discussing a DNA molecule, biologists frequently refer to the individual strands as single-stranded DNA and to the double helix as double-stranded DNA or duplex DNA.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Each DNA strand has a polarity, or directionality, like a chain of circus elephants linked trunk to tail. In this analogy, each elephant corresponds to one nucleotide along the DNA strand. The polarity is determined by the direction in which the nucleotides are pointing. The “trunk” end of the strand is called the 3′ end of the strand, and the “tail” end is called the 5′ end. In double-stranded DNA, the paired strands are oriented in opposite directions, with the 5′ end of one strand aligned with the 3′ end of the other strand. The molecular basis of the polarity, and the reason for the opposite orientation of the strands in duplex DNA, are explained in the DNA Structure and Genetic Variation chapter. In illustrating DNA molecules in this text, we use an arrow-like ribbon to represent the backbone, and we use tabs jutting off the ribbon to represent the nucleotides. The polarity of a DNA strand is indicated by the direction of the arrow-like ribbon. The tail of the arrow represents the 5′ end of the DNA strand, the head the 3′ end.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Beyond the most optimistic hopes, knowledge of the structure of DNA immediately gave clues to its function: 1. The sequence of bases in DNA could be copied by using each of the separate “partner” strands as a pattern for the creation of a new partner strand with a complementary sequence of bases. 2. The DNA could contain genetic information in coded form in the sequence of bases, analogous to letters printed on a strip of paper. 3. Changes in genetic information (mutations) could result from errors in copying in which the base sequence of the DNA became altered. In the remainder of this chapter, we discuss some of the implications of these clues.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

An Overview of DNA Replication Watson and Crick noted that the structure of DNA itself suggested a mechanism for its replication. “It has not escaped our notice,” they wrote, “that the specific base pairing we have postulated immediately suggests a copying mechanism.” The copying process in which a single DNA molecule gives rise to two identical molecules

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

is called replication. The replication mechanism that Watson and Crick had in mind is illustrated in FIGURE 1.7.



Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.7 Replication of DNA. (A) Replication of a DNA duplex as originally envisioned by Watson and Crick. As the parental strands separate, each parental strand serves as a template for the formation of a new daughter strand by means of A–T and G–C base pairing. (B) Greater detail showing how each of the parental strands serves as a template for the production of a complementary daughter strand, which grows in length by the successive addition of single nucleotides to the 3′ end.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

As shown in part A of Figure 1.7, the strands of the original (parent) duplex separate, and each individual strand serves as a pattern, or template, for the synthesis of a new strand (replica). The replica strands are synthesized by the addition of successive nucleotides in such a way that each base in the replica is complementary (in the Watson–Crick pairing sense) to the base across the way in the template strand (Figure 1.7B). Although the mechanism in Figure 1.7 is simple in principle, it is a complex process that is fraught with geometrical problems and requires a

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

variety of enzymes and other proteins. The details are examined in the DNA Replication and Sequencing chapter. The end result of replication is that a single double-stranded molecule becomes replicated into two copies with identical sequences:

Here the bases in the newly synthesized strands are shown in red. In the duplex on the left, the top strand is the template from the parental molecule and the bottom strand is newly synthesized; in the duplex on the right, the bottom strand is the template from the parental molecule and the top strand is newly synthesized. Note in Figure 1.7B that in the synthesis of each new strand, new nucleotides are added only to the 3′ end of the growing chain: The obligatory elongation of a DNA strand only at the 3′ end is an essential feature of DNA replication. This was a point not recognized by Watson and Crick in 1953, but rather one that became clear once the cellular machinery of DNA replication was characterized.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

SUMMING UP ■ DNA consists of a double helix, with the two strands held together by complementary base pairing. ■ The two strands of the double helix have opposite polarities.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Each strand of a double-helical DNA molecule serves as a template for synthesis of a new one during DNA replication.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.3 Genes and Proteins The structure of DNA, by itself, suggested how two of two of the four necessary properties of the genetic material could be explained. First, complementary base pairing provides a mechanism for replication (Figure 1.7). Second, if the sequence of nucleotides along the DNA is thought of as a string of letters on a sheet of paper, then the genes could be envisioned as distinct

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

words that form sentences and paragraphs that give meaning to the pattern of letters. Based on the structure of DNA alone, however, it is not clear what is coded for. As we will see, a contemporaneous series of experiments showed that what is encoded are proteins, a class of macromolecules that carry out most of the biochemical activities in the cell. Cells are largely made up of proteins. These include structural proteins that give the cell rigidity and mobility, proteins that form pores in the cell membrane to control the traffic of small molecules into and out of the cell, and receptor proteins that regulate cellular activities in response to molecular signals from the growth medium or from other cells. Proteins are also responsible for most of the metabolic activities of cells. They are essential for the synthesis and breakdown of organic molecules and for generating the chemical energy needed for cellular activities. In 1878, the term enzyme was introduced to refer to the biological catalysts that accelerate biochemical reactions in cells. By 1900, thanks largely to the work of the German biochemist Emil Fischer, enzymes were shown to be proteins. As often happens in science, nature's “mistakes” provide clues as to how things work. Such was the case in establishing a relationship between genes and disease, because a “mistake” in a gene (a mutation) can result in a “mistake” (lack of function) in the corresponding protein. This provided a fruitful avenue of research for the study of genetics.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Inborn Errors of Metabolism as a Cause of Hereditary Disease More than a century ago, the British physician Archibald Garrod realized that certain heritable diseases followed the rules of transmission that Mendel had described for his garden peas. In 1908, Garrod gave a series of lectures in which he proposed a fundamental hypothesis about the relationship between heredity, enzymes, and disease: Any hereditary disease in which cellular metabolism is abnormal results from an inherited defect in an enzyme. Such diseases became known as inborn errors of metabolism, a term still in use today.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Garrod studied a number of inborn errors of metabolism in which the patients excreted abnormal substances in their urine. One of these diseases was alkaptonuria. In this case, the abnormal substance excreted is homogentisic acid:

An early name for homogentisic acid was alkapton—hence the name alkaptonuria for this disease. Even though alkaptonuria is rare, with an incidence of about one in 200,000 people, it was well known even before Garrod studied it. The disease itself is relatively mild, but it has one striking symptom: The urine of the patient turns black because of the oxidation of homogentisic acid (FIGURE 1.8).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

For this reason, alkaptonuria is also called black urine disease. An early case was described in the year 1649: The patient was a boy who passed black urine and who, at the age of fourteen years, was submitted to a drastic course of treatment that had for its aim the subduing of the fiery heat of his viscera, which was supposed to bring about the condition in question by charring and blackening his bile. Among the measures

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

prescribed were bleedings, purgation, baths, a cold and watery diet, and drugs galore. None of these had any obvious effect, and eventually the patient, who tired of the futile and superfluous therapy, resolved to let things take their natural course. None of the predicted evils ensued. He married, begat a large family, and lived a long and healthy life, always passing urine black as ink. (Recounted by Garrod, 1908.)

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

FIGURE 1.8 Urine from a person with alkaptonuria turns black because of the oxidation of the homogentisic acid that it contains.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Courtesy of Daniel De Aguiar.

Garrod was primarily interested in the biochemistry of alkaptonuria, but he took note of family studies that indicated that the disease was inherited as though it were due to a defect in a single gene. As to the biochemistry, he deduced that the problem in alkaptonuria was the patients' inability to break down the phenyl ring of six carbons that is present in homogentisic acid. Where does this ring come from? Most animals obtain it from foods in their diet. Garrod proposed that homogentisic acid originates as a breakdown product of two amino acids, phenylalanine and tyrosine, which also contain a phenyl ring. An amino acid is one of the “building blocks” from which proteins are made. Phenylalanine and tyrosine are constituents of normal proteins. The scheme that illustrates the relationship between the molecules is shown in FIGURE 1.9. Any such sequence of biochemical reactions is called a biochemical pathway or a metabolic pathway. Each arrow in the pathway

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

represents a single step depicting the transition from the “input” or substrate molecule, shown at the head of the arrow, to the “output” or product molecule, shown at the tip. Biochemical pathways are usually oriented either vertically with the arrows pointing down, as in Figure 1.9, or horizontally, with the arrows pointing from left to right. Garrod did not know all of the details of the pathway in Figure 1.9, but he did understand that the key step in the breakdown of homogentisic acid is the breaking open of the phenyl ring and that the phenyl ring in homogentisic acid comes

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

from dietary phenylalanine and tyrosine.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.9 Metabolic pathway for the breakdown of phenylalanine and tyrosine. Each step in the pathway, represented

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

by an arrow, requires a specific enzyme to catalyze the reaction. The key step in the breakdown of homogentisic acid is the breaking open of the phenyl ring.

What allows each step in a biochemical pathway to occur? Garrod correctly surmised that each step requires a specific enzyme to catalyze the reaction for the chemical transformation. Persons with an inborn error of metabolism, such as alkaptonuria, have a defect

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

in a single step of a metabolic pathway because they lack a functional enzyme for that step. When an enzyme in a pathway is defective, the pathway is said to have a block at that step. One frequent result of a blocked pathway is that the substrate of the defective enzyme accumulates. Observing the accumulation of homogentisic acid in patients with alkaptonuria, Garrod proposed that there must be an enzyme whose function is to open the phenyl ring of homogentisic acid and that this enzyme is missing in these patients. Isolation of the enzyme that opens the phenyl ring of homogentisic acid was not actually achieved until 50 years after Garrod's lectures. In the average person, it is found in cells of the liver, and just as Garrod had predicted, the enzyme is defective in patients with alkaptonuria. The pathway for the breakdown of phenylalanine and tyrosine, as it is understood today, is shown in FIGURE 1.10. In this figure, the emphasis is on the enzymes rather than on the structures of the metabolites, or small molecules, on which the enzymes act. Each step in the pathway requires the presence of a particular enzyme that catalyzes that step.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.10 Inborn errors of metabolism that affect the breakdown of phenylalanine and tyrosine. An inherited disease results when any of the enzymes is missing or defective. Alkaptonuria results from a mutant homogentisic acid 1,2dioxygenase; phenylketonuria results from a mutant phenylalanine hydroxylase.

Although Garrod knew only about alkaptonuria, in which the defective enzyme is homogentisic acid 1,2-dioxygenase, we now know the clinical consequences of defects in the other enzymes. Unlike alkaptonuria, which is a relatively benign inherited disease, the others are very serious. The condition known as phenylketonuria (PKU) results from the absence of (or a defect in) the enzyme phenylalanine hydroxylase (PAH). When this step in the pathway is blocked, phenylalanine accumulates. The excess phenylalanine is broken down into harmful metabolites that cause

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

defects in myelin formation that damage a child's developing nervous system and lead to severe intellectual disability.

ROOTS OF DISCOVERY The Black Urine Disease Archibald E. Garrod (1908) St. Bartholomew's Hospital London, England

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Inborn Errors of Metabolism Although he was a distinguished physician, Garrod's lectures on the relationship between heredity and congenital defects in metabolism had no impact when they were delivered. The important concept that one gene corresponds to one enzyme (the “one gene–one enzyme hypothesis”) was developed independently in the 1940s by George W. Beadle and Edward L. Tatum, who used the bread mold Neurospora crassa as their experimental organism. When Beadle finally became aware of Garrod's hypothesis about inborn errors of metabolism, he was generous in praising it. This excerpt shows Garrod at his best, interweaving history, clinical medicine, heredity, and biochemistry in his account of alkaptonuria. The excerpt also illustrates how the severity of a genetic disease depends on its social context. Garrod writes as though alkaptonuria were a harmless curiosity. This is indeed largely true when the life expectancy is short. With today's longer life span, alkaptonuria patients accumulate the dark pigment in their cartilage and joints and can eventually develop arthritis.

“We may further conceive that the splitting of the phenyl ring in normal metabolism is the work of a special enzyme and that in

”

congenital alkaptonuria this enzyme is wanting.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Garrod's book Inborn Errors of Metabolism, published in 1903, dealt with a number of inherited metabolic disorders, but two chapters (four and five) were devoted to alkaptonuria. In them, he proposes two critical hypotheses. The first hypothesis relates to the source of the homogentisic acid in the urine of affected individuals. Two explanations are possible [for] the fact that alcaptonurics excrete homogentisic acid whereas normal persons do not. Either the alcapton acid is a strictly abnormal product formed by a perverted metabolism of tyrosin and phenylalanin, or it is an intermediate product of normal metabolism, which in alcaptonurics escapes further change … It appears to me that the evidence in favour of the theory of an intermediate product far outweighs that which can be brought against it (p. 38) …

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The second hypothesis is the one that proved to be a cornerstone of modern genetics, connecting Mendel's laws with metabolic traits. It was pointed out [by others] that the mode of incidence of alkaptonuria finds a ready explanation if the anomaly be regarded as a rare recessive character in the Mendelian sense … Of the cases of alkaptonuria, a very large proportion have been in the children of first cousin marriages … It is also noteworthy that, if one takes families with five or more children [with both parents normal and at least one child affected with alkaptonuria], the totals work out in strict conformity to Mendel's law, i.e., 57 [normal children]:19 [affected children] in the proportions 3:1 (pp. 23–25). It is completely understandable why Beadle had high praise for Garrod. In fact, the hypotheses in Inborn Errors of Metabolism were in essence those that Beadle and Tatum tested in their experiments with Neurospora. Remarkably, Garrod arrived at

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

them through synthesis of medical and physiological observations (his own and that of others) that were not made in the context of genetic analysis. Indeed, in 1903, that context was in its infancy. Source: Harris, H. (1963). Garrod, A. (1903) Inborn errors of metabolism. Oxford University Press.

However, if PKU is diagnosed in children soon enough after birth, they can be placed on a specially formulated diet that is low in phenylalanine. The child is allowed only as much phenylalanine as can be used in the synthesis of proteins, thereby ensuring excess phenylalanine does not accumulate. The special diet is very strict. It excludes meat, poultry, fish, eggs, milk and milk products, legumes, nuts, and bakery goods manufactured with regular flour. These foods are replaced by an expensive synthetic formula. With the special diet, however, the detrimental effects of excess phenylalanine on intellectual development can largely be avoided, although in adult women with PKU who are pregnant, the fetus is at risk.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In many countries, including the United States, all newborn babies have their blood tested for chemical signs of PKU. Routine screening is cost-effective because PKU is relatively common. In the United States, the incidence is about 1 in 8000 among Caucasian births. The disease is less common in other ethnic groups. In the metabolic pathway depicted in Figure 1.10, defects in the breakdown of tyrosine or of 4-hydroxyphenylpyruvic acid lead to types of tyrosinemia. Like PKU, these diseases have severe effects: Type II is associated with skin lesions and intellectual disability, and Type III with severe liver dysfunction.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

The genes for the enzymes in the biochemical pathway in Figure 1.10 have all been identified and the nucleotide sequence of the DNA determined. In the following list, and throughout this text, we use the standard typographical convention that genes are written in italic type, whereas gene products are not printed in italics. This convention is convenient, because it means that the protein product of a gene can be represented with the same symbol as the gene itself, but whereas the gene symbol is in italics, the protein symbol is not. In Figure 1.10, the numbers correspond to the following genes and enzymes: 1. The gene PAH on the long arm of chromosome 12 encodes phenylalanine hydroxylase (PAH). 2. The gene TAT on the long arm of chromosome 16 encodes tyrosine aminotransferase (TAT). 3. The gene HPD on the long arm of chromosome 12 encodes 4-hydroxyphenylpyruvic acid dioxygenase (HPD). 4. The gene HGD on the long arm of chromosome 3 encodes homogentisic acid 1,2 dioxygenase (HGD).

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

SUMMING UP ■ Archbold Garrod identified the genetic basis of “inborn errors of metabolism.” ■ Diseases such as alkaptonuria are the result of the absence of a particular enzyme in a biochemical pathway. ■ With some inborn errors of metabolism, such as phenylketonuria, dietary regulation can prevent adverse symptoms from developing.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.4 Genetic Analysis The genetic implications of Garrod's research on inborn errors of

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

metabolism were not widely appreciated, most likely because his writings focused primarily on biochemical pathways rather than inheritance. And, of course, since the genes and enzymes were being studied in humans, the potential for classical genetic analysis was limited. The definitive connection between genes and enzymes came from studies carried out in the 1940s by George W. Beadle and Edward L. Tatum using a filamentous fungus Neurospora crassa, commonly called red bread mold—an organism they chose because both genetic and biochemical analyses could be done with ease. In these experiments, Beadle and Tatum identified new mutations, each of which caused a block in the metabolic pathway for the synthesis of some needed nutrient, and showed that each of these blocks corresponded to a defective enzyme needed for one step in the pathway. Their research was important not only because it solidified the link between genetics and biochemistry, but also because their experimental approach, now called genetic analysis, has been successfully applied to understanding a wide range of biological processes, from the genetic control of the cell cycle and cancer to the genetic influences on development and behavior. For this reason, we will examine the Beadle—Tatum experiments in some detail.

ROOTS OF DISCOVERY One Gene, One Enzyme George W. Beadle and Edward L. Tatum (1941) Stanford University, Stanford, California

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Genetic Control of Biochemical Reactions in Neurospora How do genes control metabolic processes? The suggestion that genes control enzymes was made very early in the history of genetics, most notably by the British physician Archibald

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Garrod in his 1903 book Inborn Errors of Metabolism. Nevertheless, the precise relationship between genes and enzymes was still uncertain. Perhaps each enzyme is controlled by more than one gene, or perhaps each gene contributes to the control of several enzymes. The classic experiments of Beadle and Tatum showed that the relationship is usually remarkably simple: One gene codes for one enzyme. Their pioneering experiments united genetics and biochemistry, and for the “one gene, one enzyme” concept, Beadle and Tatum were awarded a Nobel Prize in 1958 (Joshua Lederberg shared the prize for his contributions to microbial genetics). Because we now know that some enzymes contain polypeptide chains encoded by two (or occasionally more) different genes, a more accurate statement of the principle is “one gene, one polypeptide.” Beadle and Tatum's experiments also demonstrate the importance of choosing the right organism. Neurospora had been introduced as a genetic organism only a few years earlier, and Beadle and Tatum realized that they could take advantage of this organism's ability to grow on a simple medium composed of known substances.

“These preliminary results appear to us to indicate that the approach may offer considerable promise as a method of learning more about how genes regulate development and

”

function.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Beadle and Tatum's work was published in 1941 and can be found at the reference at the end of this feature. In it, they point out the limitations of starting with the physiological basis of a trait (such as black urine disease) and attempting to determine its genetic basis. First, these analyses are limited to traits whose variants are nonlethal. Second, the variants must have visible effects. To get around these problems, Beadle and Tatum turned the problem on its head.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

[These limitations] have led us to investigate the general problem of the genetic control of development and metabolic reactions by reversing the ordinary procedure … [by setting out] to determine if and how genes control known biochemical reactions … If the organism must be able to carry out a certain chemical reaction to survive on a given medium, a mutant unable to do this will obviously be lethal on this medium…. [It can be] studied, however, if it will grow on a medium to which has been added the essential product of the genetically blocked reaction…. Thus, rather than starting with observed differences in traits among individuals, Beadle and Tatum started by generating mutations (in their case, mutations resulting from X-irradiation of Neurospora cells) and then identified those that on minimal medium are lethal, but on medium supplemented with the normal product of the mutated gene. This experimental approach ranks among the most important experimental tool of genetic analysis. Source: G. W. Beadle and E. L. Tatum, Genetic Control of Biochemical Reactions in Neurospora. Proc. Natl. Acad. Sci. USA 27 (1941): 499–506.

Mutant Genes and Defective Proteins

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

N. crassa grows in the form of filaments on a great variety of substrates, including a laboratory medium containing only inorganic salts, a sugar, and the vitamin biotin. Such a minimal medium contains only the nutrients that are essential for growth of the organism. The N. crassa filaments consist of a mass of branched threads separated into interconnected, multinucleate compartments, which allow free interchange of nuclei and cytoplasm. Each nucleus contains a single set of seven chromosomes. Beadle and Tatum recognized that the ability of

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Neurospora to grow in minimal medium implied that the organism must be able to synthesize all metabolic components other than biotin. If the biosynthetic pathways needed for growth are controlled by genes, then a mutation in a gene responsible for synthesizing an essential nutrient would be expected to render a strain unable to grow unless the strain was provided with the nutrient. These ideas were tested in the following way. Spores of nonmutant Neurospora were irradiated with either x-rays or ultraviolet light to produce mutant strains with various nutritional requirements. (Why these treatments cause mutations is discussed in the Mutation, Repair, and Recombination chapter.) The isolation of a set of mutants affecting any biological process—in this case, metabolism —is called a mutant screen. In the initial step for identifying mutants, summarized in FIGURE 1.11, the irradiated spores (purple) were used in crosses with an untreated strain (green). Ascospores produced by the sexual cycle in fruiting bodies were individually germinated in a complete medium—that is, a complex medium enriched with a variety of amino acids, vitamins, and other substances expected to be essential metabolites whose synthesis could be blocked by a mutation. Even those ascospores containing a new mutation affecting synthesis of an essential nutrient would be expected to germinate and grow in complete medium.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.11 Beadle and Tatum obtained mutants of the filamentous fungus Neurospora crassa by exposing asexual spores to x-rays or ultraviolet light. The treated spores were used to start the sexual cycle in fruiting bodies. After any pair of cells and their nuclei undergo fusion, meiosis takes place almost immediately and results in eight sexual spores (ascospores) included in a single ascus. These are removed individually and cultured in complete medium. Ascospores that carry new nutritional mutants are identified later by their inability to grow in minimal medium.

To identify which of the irradiated ascospores contained a new mutation affecting the synthesis of an essential nutrient, conidia from each culture were transferred to minimal medium (FIGURE 1.12A). The vast majority of cultures could grow on minimal medium; these were discarded because they lacked any new mutation of the desired type. The retained cultures were the small number unable to grow in minimal medium, because they contained a new mutation blocking the synthesis of some essential nutrient.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.12 (A) Mutant spores can grow in complete medium but not in minimal medium. (B) Each new mutant is tested for growth in minimal medium supplemented with a mixture of nutrients. (C) Mutants that can grow on minimal medium supplemented with amino acid are tested with each amino acid individually. (D) Mutants unable to grow in the absence of arginine are tested with likely precursors of arginine.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Samples from a portion of each mutant culture were then transferred to a series of media to determine whether the mutation resulted in a requirement for a vitamin, an amino acid, or some other substance (FIGURE 1.12B). In the example illustrated, the mutant strain requires one (or possibly more than one) amino acid, because a mixture of all amino acids added to the minimal medium allows growth. Because the proportion of irradiated cultures with new mutations was very small, only a negligible number of cultures contained two or more new mutations that had occurred

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

simultaneously. For nutritional mutants requiring amino acids, further experiments testing each amino acid individually usually revealed that only one additional amino acid was needed in the minimal medium to support growth. In FIGURE 1.12C, the mutant strain requires the amino acid arginine. As early as the 1940s, some of the possible intermediates in amino acid biosynthesis had been identified. They were recognized by their chemical resemblance to the amino acid and by their presence at low levels in the cells of organisms. In the case of arginine, two candidates were ornithine and citrulline. All mutants requiring arginine, therefore, were tested in medium supplemented with either ornithine alone or citrulline alone (FIGURE 1.12D). One class of arginine-requiring mutants, designated Class I, was able to grow in minimal medium supplemented with either ornithine, citrulline, or arginine. Other mutants, designated Class II, were able to grow in minimal medium supplemented with either citrulline or arginine, but not ornithine. A third class, Class III, was able to grow only in minimal medium supplemented with arginine. The types of arginine-requiring mutants illustrate the logic of genetic analysis as applied to metabolic pathways. The logic is easiest to see in the context of the metabolic pathway, shown in FIGURE 1.13, where arginine is the end product of a linear

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

metabolic pathway starting with some precursor metabolite, and ornithine and citrulline are intermediates in the pathway. The mutants support this structure of the pathway for the following reasons: ■ Mutants able to grow in the presence of either ornithine, citrulline, or arginine must have a metabolic block between the precursor metabolite and both of the intermediates. ■ Mutants able to grow only in the presence of arginine must have a metabolic block in the pathway between arginine and the most “downstream” of the intermediates.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Mutants able to grow in the presence of citrulline or arginine but not ornithine imply that ornithine is “upstream” from citrulline in the metabolic pathway.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.13 Metabolic pathway for arginine biosynthesis inferred from genetic analysis of Neurospora mutants.

The structure of the pathway was further confirmed by the observations that Class III mutants accumulate citrulline and Class II mutants accumulate ornithine. Finally, it was shown by direct biochemical experiment that the inferred enzymes were actually present in nonmutant strains but either absent or nonfunctional in mutant strains.

Complementation Test for Mutations in the Same Gene Beadle and Tatum were fortunate to study metabolic pathways in a relatively simple organism in which one gene corresponds to one

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

enzyme. In such a situation, genetic analysis of the mutants reveals a great deal more about the metabolic pathway than merely the order of the intermediates. When each mutation is classified according to the particular gene it is in, and when all the mutations in each gene are grouped together, each set of mutations—and therefore each individual gene—corresponds to one enzymatic step in the metabolic pathway. In Figure 1.13, for example, the mutants in Class I each correspond to one of three genes, which implies that there are three steps in the pathway between the precursor

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

and ornithine. Similarly, the mutants in Class III each correspond to one of two genes, which implies that there are two steps in the pathway between citrulline and arginine. However, all of the mutants in Class II have mutations in the same gene, which implies only one step in the pathway between ornithine and citrulline. Mutations that have defects in the same gene are identified by means of a complementation test, in which two mutations are brought together into the same cell. In most multicellular organisms and even some sexual unicellular organisms, the usual way to do this is by means of a mating. When two parents, each carrying one of the two mutations, are crossed, fertilization brings the reproductive cells containing the two mutations together. As the result of ordinary cell division, each cell in the offspring then carries one copy of each mutant gene. In Neurospora, this procedure does not work because nuclear fusion is followed almost immediately by the formation of ascospores, each of which has only one set of chromosomes. Complementation tests are nevertheless possible in Neurospora owing to the multinucleate nature of the filaments. Certain strains, including those studied by Beadle and Tatum, have the property that when the filaments from two mutant (or nonmutant) organisms come into physical contact, the filaments fuse and the new filament

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

contains multiple nuclei from each of the participating partners. This sort of hybrid filament, which is called a heterokaryon, contains mutant forms of both genes. The word roots of the term heterokaryon mean “different nuclei.” (A list of the most common word roots used in genetics can be found at the end of this text.) When a heterokaryon formed from two nutritional mutants is inoculated into minimal medium, it may either grow or fail to grow. If it grows in minimal medium, the mutant genes are said to undergo complementation, and this result indicates that the mutations are in different genes. Conversely, if the heterokaryon fails to grow in minimal medium, the result indicates noncomplementation, and the two mutations are inferred to be in the same gene.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

To illustrate this concept, let's consider two hypothetical examples: 1. Heterokaryons are formed between Class II and Class III mutants. In this case, the nuclei containing the Class II mutants are unable to convert ornithine to citrulline, but they do produce the enzyme (or enzymes) necessary to convert citrulline to arginine. The Class III mutants, in contrast, can perform the ornithine-to-citrulline conversion, but they are unable to convert citrulline to arginine. In the same heterokaryon, therefore, both steps can occur, so that growth on minimal medium occurs. These two mutations are said to be complementary. 2. Heterokaryons are formed between two independently derived Class II mutations. Neither of these mutations can carry out the ornithine-to-citrulline step; hence no growth on minimal medium will occur. These mutations are noncomplementary.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Now let's consider two Class III mutations. Are they complementary? If the citrulline-to-arginine conversion includes only a single biochemical step, then we would expect them to be noncomplementary. Beadle and Tatum, however, discovered that some pairs of mutants did complement each other, whereas others did not. The theory that one gene codes for one enzyme led them to conclude that more than one enzyme is required for this step. The inferences from complementation or noncomplementation

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

emerge from the logic illustrated in FIGURE 1.14. Here the multinucleate filament is shown, and the mutant nuclei are color coded according to which of two different genes (red or purple) is mutant. The red and purple lines represent the proteins encoded in the mutant nuclei, and an asterisk represents a defect in the protein at that position resulting from a mutation in the corresponding gene.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.14 Molecular interpretation of a complementation test using heterokaryons to determine whether two mutant strains have mutations in different genes (A) or mutations in the same gene (B). In (A), each nucleus contributes a nonmutant form of one or the other polypeptide chain, so the heterokaryon is able to grow in minimal medium. In (B), both nuclei contribute a mutant form of the same polypeptide chain; hence, no nonmutant form of that polypeptide can be synthesized and the heterokaryon is unable to grow in minimal medium.

Part A depicts the situation in which the mutant strains have mutations in different genes. In the heterokaryon, the red nuclei produce mutant forms of the red protein and normal forms of the purple protein, whereas the purple nuclei produce mutant forms of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

the purple protein and normal forms of the red protein. The result is that the red/purple heterokaryon has normal forms of both the red and purple protein. It also has mutant forms of both proteins, but these do not matter. What matters is that the normal forms allow the heterokaryon to grow on minimal medium because all needed nutrients can be synthesized. In other words, the normal purple gene in the red nucleus complements the defective purple gene in the purple nucleus, and vice versa. The logic of complementation is captured in the ancient nursery rhyme, “Jack Sprat could eat no fat / His wife could eat no lean / And so between the two of them / They licked the platter clean,” because each partner makes up for the defect in the other.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Part B in Figure 1.14 shows a heterokaryon formed between mutants with defects in the same gene—in this case, purple. Both of the purple nuclei encode a normal form of the red protein, but each purple nucleus encodes a defective purple protein. When the nuclei are together, two different mutant forms of the purple protein are produced, so the biosynthetic pathway that requires the purple protein is still blocked and the heterokaryon is unable to grow in minimal medium. In other words, the mutants 2 and 3 in Figure 1.14 fail to complement, so they are judged to have mutations in the same gene. The following principle underlies the complementation test:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Principle of Complementation: A complementation test brings two mutant genes together in the same cell or organism. If this cell or organism is nonmutant, the mutations are said to complement each other, and the parental strains must have mutations in different genes. If the cell or organism is mutant, the mutations fail to complement each other, and the parental mutations must be in the same gene. These latter mutations form a complementation group.

Analysis of Complementation Data

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In the mutant screen for Neurospora mutants requiring arginine, Beadle and Tatum found that mutants in different classes (Class I, Class II, and Class III in Figure 1.13) always complemented one another. This result makes sense because the genes in each class encode enzymes that act at different levels between the known intermediates. However, some of the mutants in Class I failed to complement others in Class I, and some in Class III failed to complement others in Class III. These results allow the number of genes in each class to be identified. To illustrate this aspect of genetic analysis, we will consider six mutant strains in Class III. These strains were taken in pairs to form heterokaryons and their growth on minimal medium assessed. The data are shown in FIGURE 1.15. The mutant genes in the six strains are denoted x1, x2, and so forth, and the data are presented in the form of a matrix in which a plus sign (+) indicates growth in minimal medium (complementation) and a minus sign (−) indicates lack of growth in minimal medium (lack of complementation). The diagonal entries are all minus signs, which reflects the fact that two copies of the identical mutation cannot show complementation. The pattern of + and − signs in the matrix

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

indicates that mutations x1 and x5 fail to complement each other; hence x1 and x5 are mutations in the same gene. Likewise, mutations x2, x3, x4, and x6 fail to complement one another in all pairwise combinations; hence x2, x3, x4, and x6 are all mutations in the same gene, but a different gene from that represented by x1

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

and x5.

FIGURE 1.15 Interpreting the results of complementation tests. Rows and columns are particular, independently derived mutations. Pluses indicate pairs that complement; minuses are ones that do not. A complementation group is defined as all mutations that do not complement one another but do complement all other mutations. In this case, x1 and x4 constitute one group, while x2, x3, x4, and x6 make up a second group.

Each of the groups of noncomplementary mutations is called a complementation group. As we have seen, each complementation group defines a gene, so the complementation test actually provides the geneticist's operational definition:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

A gene is defined experimentally as a set of mutations that make up a single complementation group. Any pair of mutations within a complementation group fails to complement each other. The mutations in Figure 1.15 therefore represent two genes, and mutation of any one of them results in the inability of the strain to convert citrulline to arginine. On the assumption that one gene encodes one enzyme, which is largely true for metabolic enzymes in Neurospora, the pathway from citrulline to arginine depicted in Figure 1.13 must consist of two steps, with an unknown intermediate in between the steps. This intermediate was later found to be argininosuccinate. Likewise, the Class I mutants define three complementation groups, so there are three enzymatic steps from the precursor to ornithine. These intermediates also were soon identified. Finally, Class II mutations all fail to complement one another, and the finding of only one complementation group means that there is but a single enzymatic step that converts ornithine to citrulline.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Other Applications of Genetic Analysis The type of genetic analysis pioneered by Beadle and Tatum is immensely powerful for identifying the genetic control of complex biological processes. Their approach lays out a systematic path—a cookbook if you will—for gene discovery, one that we will see used repeatedly throughout this text: 1. Decide which process you want to study, and figure out which characteristics would be displayed by mutant organisms with a disruption in that process.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

2. Perform mutant screening for mutants showing these characteristics. 3. Carry out complementation tests to find out how many different genes you have identified. 4. Identify the products of those genes, and determine what they do, how they interact with each other, and in which order they function. Beadle and Tatum analyzed many metabolic pathways for a wide

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

variety of essential nutrients, but their experiments were especially important in deciphering the pathways of amino acid biosynthesis. Their findings over the span of just a few years were remarkable, in that they deduced much more about the nature of biosynthetic pathways than had been learned from decades of biochemical research. Beadle and Tatum were awarded the 1958 Nobel Prize in Physiology or Medicine for their research, and in the intervening years many more Nobel Prizes in Physiology or Medicine were awarded in which genetic analysis carried out along the lines of Beadle and Tatum played a significant role. Here is a list, with quotations from the official citations of the Nobel Foundation: ■ 1958: George Beadle and Edward Tatum “for their discovery that genes act by regulating definite chemical events,” shared with Joshua Lederberg “for his discoveries concerning genetic recombination and the organization of the genetic material of bacteria.” ■ 1965: François Jacob, André Lwoff, and Jacques Monod “for their discoveries concerning genetic control of enzyme and virus synthesis.” ■ 1995: Edward B. Lewis, Christiane Nüsslein-Volhard, and Eric F. Wieschaus “for their discoveries concerning the genetic control of early embryonic development.”

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

■ 2001: Leland H. Hartwell, Tim Hunt, and Sir Paul Nurse “for their discoveries of key regulators of the cell cycle.” ■ 2002: Sydney Brenner, H. Robert Horvitz, and John E. Sulston “for their discoveries concerning genetic regulation of organ development and programmed cell death.” ■ 2007: Mario R. Capecchi, Martin J. Evans, and Oliver Smithies “for their discoveries of principles for introducing specific gene modifications in mice by the use of embryonic stem cells.” ■ 2009: Elizabeth Blackburn, Carol Greider, and Jack Szostak “for the discovery of how chromosomes are protected by telomeres and the enzyme telomerase.” ■ 2013: James Rothman, Randy Shekman, and Thomas Sudhof “for their discoveries of machinery regulating vesicle traffic, a major transport system in our cells.”

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ 2015: Tomas Lindahl, Paul Modrich, and Aziz Sancar “for mechanistic studies of DNA repair.” The Beadle and Tatum experiments established that a defective enzyme results from a mutant gene, but how? For all they knew, genes were enzymes. This would have been a logical hypothesis at the time. We now know that the relationship between genes and enzymes is somewhat indirect. With a few exceptions, each enzyme is encoded in a particular sequence of nucleotides present in a region of DNA. The DNA region that codes for the enzyme, as well as adjacent regions that regulate when and in which cells the enzyme is produced, make up the “gene” that encodes the enzyme. Next, we turn to the issue of how genes code for enzymes and other proteins.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

SUMMING UP ■ Genetic analysis is an experimental methodology whereby inferences based on the outcomes of genetic crosses can be used to elucidate a wide variety of biological processes. ■ Using Neurospora crassa, Beadle and Tatum developed the methods of mutant screening. ■ Complementation analysis is a powerful means for determining the number of steps in a biochemical pathway.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ These analyses led Beadle and Tatum to the “one gene, one enzyme” hypothesis.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.5 Gene Expression: The Central Dogma Watson and Crick were correct in proposing that the genetic information in DNA is contained in the sequence of bases in a manner analogous to letters printed on a strip of paper. In a region of DNA that directs the synthesis of a protein, the genetic code for the protein is contained in only one strand, and it is decoded in a linear order. A typical protein is made up of one or more

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

polypeptide chains; each polypeptide chain consists of a linear sequence of amino acids connected end to end. For example, the enzyme PAH consists of four identical polypeptide chains, each 452 amino acids in length. In the decoding of DNA, each successive “code word” in the DNA specifies the next amino acid to be added to the polypeptide chain as it is being made. The amount of DNA required to code for the polypeptide chain of PAH is therefore 452 × 3 1356 nucleotide pairs. The entire gene is very much longer— about 90,000 nucleotide pairs. Only 1.5 percent of the gene is devoted to coding for the amino acids. The noncoding part includes some sequences that control the activity of the gene, but it is not known how much of the gene is involved in regulation. There are 20 different amino acids encoded by DNA. Only four bases code for these 20 amino acids, with each “word” in the genetic code consisting of three adjacent bases. For example, the base sequence ATG specifies the amino acid methionine (Met), TCC specifies serine (Ser), ACT specifies threonine (Thr), and GCG specifies alanine (Ala). There are 64 possible three-base combinations but only 20 amino acids because some combinations specify the same amino acid. For example, TCT, TCC, TCA, TCG, AGT, and AGC all code for serine (Ser), and CTT, CTC, CTA, CTG, TTA, and TTG all code for leucine (Leu).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

An example of the relationship between the nucleotide sequence in a DNA duplex and the amino acid sequence of the corresponding protein is shown in FIGURE 1.16. This particular DNA duplex is the human sequence that codes for the first seven amino acids in the polypeptide chain of PAH. The scheme outlined in Figure 1.16

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

indicates that DNA codes for protein not directly but indirectly through the processes of transcription and translation. The indirect route of information transfer is thus:

FIGURE 1.16 DNA sequence coding for the first seven amino acids in a polypeptide chain. The DNA sequence specifies the amino acid sequence through a molecule of RNA that serves as an intermediary “messenger.” Although the decoding process is indirect, the net result is that each amino acid in the polypeptide chain is specified by a group of three adjacent bases in the DNA. In this example, the polypeptide chain is that of phenylalanine hydroxylase (PAH).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

This relationship leads us to the central dogma of molecular genetics. The term dogma means “set of beliefs”; it dates from the time the idea was put forward first as a theory. Since then the “dogma” has been confirmed experimentally, but the term persists. The main concept in the central dogma, as stated by Francis Crick in 1958, is as follows:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Once “information” has passed into protein, it cannot get out again. With respect to gene expression in most cells, the central dogma can be illustrated as in FIGURE 1.17. The “information” is that which is encoded in DNA, and which specifies the sequences of proteins. Its passage from DNA to protein does not occur directly, but rather through interactions with the intermediary molecule known as ribonucleic acid (RNA). The structure of RNA is similar, but not identical, to the structure of DNA. There is a difference in the sugars in DNA and RNA (RNA contains the sugar ribose instead of deoxyribose), RNA is usually single-stranded (not a duplex), and RNA contains the base uracil (U) instead of thymine (T), which is present in DNA. Information flow between DNA and RNA follows the rules of Watson–Crick base pairing, and as we will see elsewhere in this text, in some cases it can occur in both directions (DNA → RNA or RNA → DNA). However, information flow from RNA to proteins requires translation of nucleic acid sequences into polypeptide sequences. This process is irreversible. There is no known case in which the sequence of a protein codes or the sequence of a nucleic acid (either DNA or RNA).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

FIGURE 1.17 Gene expression at the molecular level. DNA codes for RNA, and RNA codes for protein. The DNA → RNA step is transcription, and the RNA → protein step is translation.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Three types of RNA take part in the synthesis of proteins: ■ A molecule of messenger RNA (mRNA), which carries the genetic information from DNA and is used as a template for polypeptide synthesis. In most mRNA molecules, there is a high proportion of nucleotides that actually code for amino acids. For example, the mRNA for PAH is 2400 nucleotides in length and codes for a polypeptide of 452 amino acids; in this case, more than 50 percent of the length of the mRNA codes for amino acids. ■ Four types of ribosomal RNA (rRNA), which are major constituents of the cellular particles called ribosomes on which polypeptide synthesis takes place. ■ A set of approximately 45 transfer RNA (tRNA) molecules, each of which carries a particular amino acid as well as a

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

three-base recognition region that base-pairs with a group of three adjacent bases in the mRNA. As each tRNA participates in translation, its amino acid becomes the terminal subunit added to the length of the growing polypeptide chain. A tRNA that carries methionine is denoted tRNAMet, one that carries serine is denoted tRNASer, and so forth. (Because there are more than 20 different tRNAs but only 20 amino acids, some amino acids correspond to more than one tRNA.) The central dogma is the fundamental principle of molecular genetics because it summarizes how the genetic information in DNA becomes expressed in the amino acid sequence in a polypeptide chain:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The sequence of nucleotides in a gene specifies the sequence of nucleotides in a molecule of messenger RNA; in turn, the sequence of nucleotides in the messenger RNA specifies the sequence of amino acids in the polypeptide chain. Given a process as conceptually simple as DNA coding for protein, what might account for the additional complexity of RNA intermediaries? One possible explanation is that an RNA intermediate gives another level for control—for example, by degrading the mRNA for an unneeded protein. Another possible reason may be historical. RNA structure is unique in having both informational content present in its sequence of bases and a complex, folded three-dimensional structure that endows some RNA molecules with catalytic activities. Many scientists believe that in the earliest forms of life, RNA had roles both as genetic information and in catalysis. As evolution proceeded, the informational role was transferred to DNA and the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

catalytic role to protein. However, RNA became locked into its central location as a go-between in the processes of information transfer and protein synthesis. This hypothesis implies that the participation of RNA in protein synthesis is a relic of the earliest stages of evolution—a “molecular fossil.” It is supported by a variety of observations. For example, (1) DNA replication requires an RNA molecule to get started, (2) an RNA molecule is essential in the synthesis of the tips of the chromosomes, and (3) some RNA molecules act to catalyze key reactions in protein synthesis.

Transcription

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.18 depicts the manner in which genetic information is transferred from DNA to RNA. The DNA opens up, and one of the strands is used as a template for the synthesis of a complementary strand of RNA. (How the template strand is chosen is discussed in the Molecular Biology of Gene Expression chapter.) The process of making an RNA strand from a DNA template is called transcription, and the RNA molecule that is made is the transcript. The base sequence in the RNA is complementary (in the Watson–Crick pairing sense) to that in the DNA template, except that U (which pairs with A) is present in the RNA in place of T. The rules of base pairing between DNA and RNA are summarized in FIGURE 1.19.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.18 Transcription is the production of an RNA strand that is complementary in base sequence to a DNA strand. In this example, the DNA strand at the bottom is being transcribed into a strand of RNA. Note that in an RNA molecule, the base U (uracil) plays the role of T (thymine), in that it pairs with A (adenine). Each A–U pair is marked.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

FIGURE 1.19 Pairing between bases in DNA and in RNA. The DNA bases A, T, G, and C pair with the RNA bases U, A, C, and G, respectively.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Each RNA strand has a polarity—a 5′ end and a 3′ end—and, as in the synthesis of DNA, nucleotides are added only to the 3′ end of a growing RNA strand. Hence, the 5′ end of the RNA transcript is synthesized first, and transcription proceeds along the template DNA strand in the 3′-to-5′ direction. Each gene includes nucleotide sequences that initiate and terminate transcription. The RNA transcript made from any gene begins at the initiation site in the template strand, which is located “upstream” from the amino acid–coding region, and ends at the termination site, which is located “downstream” from the amino acid–coding region. For any gene, the length of the RNA transcript is very much smaller than the length of the DNA in the chromosome. For example, the transcript of the PAH gene for phenylalanine hydroxylase is about 90,000 nucleotides in length, but the DNA in chromosome 12 is about 130,000,000 nucleotide pairs. In this case, the length of the PAH transcript is less than 0.1 percent of the length of the DNA in the chromosome. A different gene in chromosome 12 would be transcribed from a different region of the DNA molecule in chromosome 12, and perhaps from the opposite strand, but the transcribed region would again be small in comparison with the total length of the DNA in the chromosome.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Translation The synthesis of a polypeptide under the direction of an mRNA molecule is known as translation. Although the sequence of bases in the mRNA codes for the sequence of amino acids in a polypeptide, the molecules that actually do the “translating” are the tRNA molecules. The mRNA molecule is translated in nonoverlapping groups of three bases called codons. For each codon in the mRNA that specifies an amino acid, there is one tRNA molecule containing a complementary group of three adjacent bases that can pair with the codon. The correct amino acid is attached to one end of the tRNA, and when the tRNA comes into line, the amino acid to which it is attached becomes the most recent addition to the growing end of the polypeptide chain. The role of tRNA in translation is illustrated in FIGURE 1.20 and can be described as follows:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The mRNA is read codon by codon. Each codon that specifies an amino acid matches with a complementary group of three adjacent bases in a single tRNA molecule. One end of the tRNA is attached to the correct amino acid, so the correct amino acid is brought into line.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.20 The role of messenger RNA in translation is to carry the information contained in a sequence of DNA bases to a ribosome, where it is translated into a polypeptide chain. Translation is mediated by transfer RNA (tRNA) molecules, each of which can base-pair with a group of three adjacent bases in the mRNA. Each tRNA also carries an amino acid. As each tRNA, in turn, is brought to the ribosome, the growing polypeptide chain is elongated.

The tRNA molecules used in translation do not line up along the mRNA simultaneously, as shown in Figure 1.20. The process of translation takes place on a ribosome, which combines with a single mRNA and moves along it from one end to the other in steps, three nucleotides at a time (codon by codon). As each new codon comes into place, the next tRNA binds with the ribosome. Then the growing end of the polypeptide chain becomes attached to the amino acid on the tRNA. In this way, each tRNA in turn serves temporarily to hold the polypeptide chain as it is being synthesized. As the polypeptide chain is transferred from each tRNA to the next in line, the tRNA that previously held the polypeptide is released from the ribosome. The polypeptide chain elongates one amino acid at each step until any one of three particular codons specifying “stop” is encountered. At this point, synthesis of the chain of amino

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

acids is finished, and the polypeptide chain is released from the ribosome. (This brief description of translation glosses over many of the details that are presented in the Molecular Biology of Gene Expression chapter.)

The Genetic Code Figure 1.20 indicates that the mRNA codon AUG specifies methionine (Met) in the polypeptide chain, UCC specifies Ser

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(serine), ACU specifies Thr (threonine), and so on. The complete decoding table is called the genetic code, and it is shown in TABLE 1.1. For any codon, the column on the left corresponds to the first nucleotide in the codon (reading from the 5′ end), the row across the top corresponds to the second nucleotide, and the column on the right corresponds to the third nucleotide. The complete codon is given in the body of the table, along with the amino acid (or translational “stop”) that the codon specifies. Each amino acid is designated by its full name and by a three-letter abbreviation as well as a single-letter abbreviation. Both types of abbreviations are used in molecular genetics.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

TABLE 1.1 The standard genetic code

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The code in Table 1.1 is the “standard” genetic code used in translation in the cells of nearly all organisms. In the Molecular Biology of Gene Expression chapter, we examine general features of the standard genetic code and the minor differences found in the genetic codes of certain organisms and cellular organelles. At this point, we are interested mainly in understanding how the genetic code is used to translate the codons in mRNA into the amino acids in a polypeptide chain. In addition to the 61 codons that code only for amino acids, there are four codons that have specialized functions: ■ The codon AUG, which specifies Met (methionine), is also the “start” codon for polypeptide synthesis. The positioning of a tRNAMet bound to AUG is one of the first steps in the initiation of polypeptide synthesis, so all polypeptide chains begin with Met. (Many polypeptides have the initial Met cleaved off after translation is complete.) In most organisms, the tRNAMet used Met

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

for initiation of translation is the same tRNAMet used to specify methionine at internal positions in a polypeptide chain. ■ The codons UAA, UAG, and UGA are each a “stop” that specifies the termination of translation and results in release of the completed polypeptide chain from the ribosome. These codons do not have tRNA molecules that recognize them, but are instead recognized by protein factors that terminate translation. How the genetic code table is used to infer the amino acid sequence of a polypeptide chain can be illustrated by using PAH again—in particular, the DNA sequence coding for amino acids 1 through 7. The DNA sequence is:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

This region is transcribed into RNA in a left-to-right direction. Because RNA grows by the addition of successive nucleotides to the 3′ end (Figure 1.18), it is the bottom strand that is transcribed. The nucleotide sequence of the RNA is that of the top strand of the DNA, except that U replaces T, so the mRNA for amino acids 1 through 7 is:

The codons are read from left to right according to the genetic code shown in Table 1.1. Codon AUG codes for Met (methionine), UCC codes for Ser (serine), and so on. Altogether, the amino acid sequence of this region of the polypeptide is:

or, in terms of the single-letter abbreviations:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

The full decoding operation for this region of the PAH gene is shown in FIGURE 1.21. In this figure, the initiation codon AUG is highlighted because some patients with PKU have a mutation in this particular codon. As might be expected from the fact that AUG is the initiation codon for polypeptide synthesis, cells in patients with this particular mutation fail to produce any of the PAH polypeptide.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Mutation and its consequences are considered next.

FIGURE 1.21 The central dogma in typical gene expression. The DNA that encodes PAH serves as a template for the production of a messenger RNA, and the mRNA serves to specify the sequence of amino acids in the PAH polypeptide chain through interactions with the ribosome and tRNA molecules. The total number of amino acids in the PAH polypeptide chain is 452, but only the first 7 are shown.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

SUMMING UP ■ Proteins consist of polypeptides—that is, chains of amino acids whose order is determined by the sequences of bases in the genes encoding them. ■ The central dogma of molecular biology states that information, in the form of molecular sequences, passes from nucleic acids to proteins, but not the reverse. ■ In gene expression, this process involves transcription of DNA into messenger RNA and translation of mRNA into protein. ■ In addition to mRNA, ribosomal RNA and transfer RNA play critical roles in the translation process.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The standard genetic code consists of 64 three-letter words, or codons, 61 of which encode amino acids.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.6 Mutation and Variation We now turn to the fourth essential property of the genetic material —namely, its ability to generate variation through mutation. The term mutation refers to any heritable change in a gene (or, more generally, in the genetic material) or to the process by which such a change takes place. One type of mutation results in a change in the sequence of bases in DNA. This kind of change may be simple,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

such as the substitution of one pair of bases in a duplex molecule for a different pair of bases. For example, a C–G pair in a duplex molecule may mutate to T–A, A–T, or G–C. The change in nucleotide sequence may also be more complex, such as the deletion or addition of base pairs. These and other types of mutations are considered in the Mutation, Repair, and Recombination chapter.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

THE WOMEN IN THE WEDDING photograph are sisters. Both have two copies of the same mutant PAH gene. The bride is the younger of the two. She was diagnosed just three days after birth and put on the PKU diet soon after. Her older sister, the maid of honor, was diagnosed too late to begin the diet and is intellectually disabled. The two-year old pictured in the photo at the right is the daughter of the married couple. They planned the pregnancy: Dietary control was strict from conception to delivery to avoid the hazards of excess phenylalanine harming the fetus. Their daughter has passed all developmental milestones with distinction.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Courtesy of Charles R. Scriver, Montreal Children's Hospital Research Institute, McGill University Health Center.

Geneticists also use the term mutant, which refers to the result of a mutation. A mutation yields a mutant gene, which in turn produces a mutant mRNA, a mutant protein, and finally a mutant organism that exhibits the effects of the mutation—for example, an inborn error of metabolism.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

DNA from patients from all over the world who have phenylketonuria has been studied to determine which types of mutations are responsible for the inborn error. A large variety of mutant types have been identified. More than 400 different mutations have been described in the gene for PAH. In some cases part of the gene is missing, so the genetic information to make a complete PAH enzyme is absent. In other cases the genetic defect is more subtle, but the result is still either the failure to produce a PAH protein or the production of an inactive PAH protein. In the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

mutation shown in FIGURE 1.22, substitution of a G–C base pair for the normal A–T base pair at the very first position in the coding sequence changes the normal codon AUG (Met) used for the initiation of translation into the codon GUG, which normally specifies valine (Val) and cannot be used as a “start” codon. The result is that translation of the PAH mRNA cannot occur, so no PAH polypeptide is made. This mutant is designated M1V because the codon for M (methionine) at amino acid position 1 in the PAH polypeptide has been changed to a codon for V (valine). Although the M1V mutant is quite rare worldwide, it is common in some localities, such as Québec Province in Canada.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 1.22 The M1V mutant in the PAH gene. The methionine codon needed for initiation mutates into a codon for valine. Translation cannot be initiated, and no PAH polypeptide is produced.

One PAH mutant that is quite common is designated R408W, which means that codon 408 in the PAH polypeptide chain has been changed from one coding for arginine (R) to one coding for tryptophan (W). This mutation is one of the four most common among European Caucasians with PKU. The molecular basis of the mutant is shown in FIGURE 1.23. In this case, the first base pair in codon 408 is changed from C–G base into T–A. As a result, the PAH mRNA has a mutant codon at position 408; specifically, it has UGG instead of CGG. Translation does occur in this mutant because everything else about the mRNA is normal, but the result is that the mutant PAH carries a tryptophan (Trp) instead of an arginine (Arg) at position 408 in the polypeptide chain. The consequence of the seemingly minor change of one amino acid is very drastic. Although the R408W polypeptide is complete, the enzyme has less than 3 percent of the activity of the normal enzyme.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

FIGURE 1.23 The R408W mutant in the PAH gene. Codon 408 for arginine (R) is mutated into a codon for tryptophan (W). The result is that position 408 in the mutant PAH polypeptide is occupied by tryptophan rather than by arginine. The mutant protein has only a low level of PAH enzyme activity.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Variation in Populations Much of the classic work in genetics, from Gregor Mendel onward, involved genetic analysis of visible differences in organismal traits that are the results of particular mutations. Most of the variants used, such as white eyes in Drosophila, shrunken kernels in maize, or even inherited diseases like alkaptonuria in humans, are quite rare. Thus, early geneticists tended to think of populations as consisting of a few rare variants among large numbers of “wild type” individuals, suggesting that overall levels of genetic variation were quite low. This picture later changed dramatically as, based on the one geneone enzyme hypothesis, it became possible to quantify genetic variation—first at the level of protein sequence, and subsequently at the level of DNA sequence. Early work with protein variation in

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

both fruit flies (Drosophila pseudoobscura) and humans yielded unexpected findings: When the protein products of several genes were examined, approximately 30 percent of them were found to be polymorphic, meaning that there were two or more variants, or alleles, for those genes present in the sampled individuals. Similar

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

findings were made for almost every species examined with these methods, leading to the recognition that variation—rather than uniformity—is the norm at the genetic level in most species. In recent years, the methods we describe in the chapter titled DNA Replication and Sequencing have allowed investigators to change their focus from single genes to entire genomes. One outcome of that effort has been the 1000 Genomes Project, an ongoing international research program seeking to survey common human genetic variation and to identify common human genetic variants associated with disease. To date, the project has sequenced 2504 genomes from 26 worldwide populations, which have been divided into 5 “superpopulations” (FIGURE 1.24). More than 80 million polymorphisms have been identified, most of which consist of single nucleotide differences (single-nucleotide polymorphisms, or SNPs). These publicly available data provide a global perspective on genetic variation in Homo sapiens, and have become an invaluable tool for researchers in medical genetics, evolutionary biology, and anthropology. Most importantly, genomic analysis of humans and other species has shown that the classical view, which suggests genetic variation consists of common wild-type alleles and rare variants, needs to be revised substantially. Much of the genome of a particular species is indeed invariant, but for significant portions of it, variation is the rule rather than the exception.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

FIGURE 1.24 Geographic distribution of populations sampled in the 1000 Genomes Project. Each dot indicates a study population; colors indicate superpopulations as indicated in the legend. Data from International Genome Sample Resource. (2016). IGSR and the 1000 Genomes Project. Retrieved from http://www.internationalgenome.org.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

SUMMING UP ■ All new genetic variation originates as a mutation, or change in DNA sequence. ■ Different, independently occurring mutations may occur in the same gene. ■ Genomic sequencing has provided powerful tools to characterize the extent and nature of genetic variation in many species, including humans.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.7 Genes and Environment Inborn errors of metabolism illustrate the general principle that genes code for proteins and mutant genes code for mutant proteins. In diseases such as PKU, mutant proteins cause such a drastic change in metabolism that a severe genetic defect results. Nevertheless, biology is not necessarily destiny—organisms are also affected by the environment. PKU serves as an example of

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

this principle, because patients who adhere to a diet restricted in the amount of phenylalanine develop intellectual capacities within the normal range. What is true in this example is true in general: Most traits are determined by the interaction of genes and environment. It is also true that most traits are affected by multiple genes. No one knows how many genes are involved in the development and maturation of the brain and nervous system, but the number must be in the thousands. This number is in addition to the genes that are required in all cells to carry out metabolism and other basic life functions. It is easy to lose sight of the multiplicity of genes when considering extreme examples, such as PKU, in which a single mutation can have such a drastic effect on intellectual development. The situation is the same as that with any complex machine. An airplane can function properly provided that thousands of parts work together in harmony, but just one defective part, if that part affects a vital system, can bring it down. Likewise, the development and functioning of every trait require a large number of genes working in harmony, but in some cases a single mutant gene can have catastrophic consequences. In short, the relationship between a gene and a trait is not necessarily a simple one. The biochemistry of organisms is a complex branching network in which different enzymes may share

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

substrates, yield the same products, or be responsive to the same regulatory elements. Most visible traits of organisms are the net result of many genes acting together and in combination with environmental factors. PKU affords examples of each of three principles governing these interactions:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1. One gene can affect more than one trait. Children with extreme forms of PKU often have blond hair and reduced body pigment. These manifestations occur because the absence of PAH is a metabolic block that prevents conversion of phenylalanine into tyrosine, which is the precursor of the pigment melanin. The relationship between severe intellectual disability and decreased pigmentation in PKU makes sense only if one knows the metabolic connections among phenylalanine, tyrosine, and melanin. If these connections were not known, the traits would seem completely unrelated. PKU is not unusual in this regard. Many mutant genes affect multiple traits through their secondary or indirect effects. The various, sometimes seemingly unrelated, effects of a mutant gene are called pleiotropic effects, and the phenomenon itself is known as pleiotropy. FIGURE 1.25 shows a cat with white fur and blue eyes, a pattern of pigmentation that is often (about 40 percent of the time) associated with deafness. Hence, deafness can be regarded as a pleiotropic effect of white coat and blue eye color. The developmental basis of this pleiotropy is unknown. 2. Any trait can be affected by more than one gene. We discussed this principle earlier in connection with the large number of genes that are required for the normal development and functioning of the brain and nervous system. Among these are genes that affect the function of the blood– brain barrier, which consists of specialized glial cells wrapped tightly around capillary walls in the brain, forming an

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

impediment to the passage of most water-soluble molecules from the blood to the brain. The blood–brain barrier therefore affects the extent to which excess free phenylalanine in the blood can enter the brain itself. Because the effectiveness of the blood–brain barrier differs among individuals, patients with

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

PKU who have very similar levels of blood phenylalanine can have dramatically different levels of cognitive development. This factor also explains in part why adherence to a controlled-phenylalanine diet is critically important in children but less so in adults; the blood–brain barrier is less well developed in children and, therefore, less effective in blocking the excess phenylalanine. Multiple genes affect even simpler metabolic traits. Phenylalanine breakdown and excretion serve as a convenient example. The metabolic pathway illustrated in Figure 1.10 indicates four enzymes play roles in this process, but even more enzymes are involved at the stage labeled “further breakdown.” Because differences in the activity of any of these enzymes can affect the rate at which phenylalanine is broken down and excreted, all of the enzymes in the pathway are important in determining the amount of excess phenylalanine in the blood of patients with PKU. 3. Most traits are affected by environmental factors as well as by genes. Here we come back to the low-phenylalanine diet. Children with PKU are not doomed to severe intellectual deficiency; their capabilities can be brought into the normal range by dietary treatment. PKU serves as an example of what motivates geneticists to try to discover the molecular basis of inherited disease. The hope is that knowing the metabolic basis of a disease will eventually lead to methods for clinical intervention through diet, medication, or other treatments that will ameliorate the severity of the disease.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

FIGURE 1.25 Among cats with white fur and blue eyes, approximately 40 percent are born deaf. The reason is that pigment cells derived from the neural crest migrate to various tissues including hair follicles, the eyes, and the middle ear, where their function is essential for hearing. Defective pigment cells resulting in white fur and blue eyes can, therefore, lead to deafness, which may be regarded as a pleiotropic effect of white fur and blue eyes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

© Ronen/Shutterstock.

SUMMING UP ■ Most traits are affected by multiple genes. ■ One gene can affect multiple traits. ■ Most traits are affected by a combination of genes and the environment.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved. Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.8 The Molecular Unity of Life The pathway for the breakdown and excretion of phenylalanine is by no means unique to human beings. One of the remarkable generalizations to have emerged from molecular genetics is that organisms that are very distinct—for example, plants and animals —share many features in their genetics and biochemistry. These similarities indicate a fundamental “unity of life”: All creatures on Earth share many features of the genetic apparatus, including genetic information encoded in the sequence of bases in DNA, transcription into RNA, and translation into protein on ribosomes with the use of transfer RNAs. All creatures also share certain characteristics in their biochemistry, including many enzymes and other proteins that are similar in amino acid sequence, three-dimensional structure, and function.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Prokarya, Archaea, and Eukarya Organisms share a common set of similar genes and proteins because they evolved by descent from a common ancestor. The process of evolution takes place when a population of organisms gradually changes in its genetic composition through time. Evolutionary changes in genes and proteins result in differences in metabolism, development, and behavior among organisms, which allows them to become progressively better adapted to their environments. From an evolutionary perspective, the unity of fundamental molecular processes in organisms alive today reflects inheritance from a distant common ancestor in which the molecular mechanisms were already in place.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Not just the unity of life but many other features of living organisms become comprehensible from an evolutionary perspective. For example, the interposition of an RNA intermediate in the basic flow of genetic information from DNA to RNA to protein makes sense if the earliest forms of life used RNA for both genetic information and enzyme catalysis. The importance of the evolutionary perspective in understanding aspects of biology that seem pointless or needlessly complex is summed up in a famous aphorism of the evolutionary biologist Theodosius Dobzhansky: “Nothing in biology makes sense except in the light of evolution.”

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Biologists distinguish three major domains of organisms: 1. Prokarya. This group includes most bacteria and cyanobacteria (formerly called blue-green algae). Cells of these organisms lack a membrane-bounded nucleus and mitochondria, are surrounded by a cell wall, and divide by binary fission. 2. Archaea. This group was initially discovered among microorganisms that produce methane gas or that live in extreme environments, such as hot springs or high salt concentrations. They are widely distributed in more normal environments as well. Superficially resembling bacteria, the cells of archaeans show important differences in the manner in which their membrane lipids are chemically linked. The machinery for DNA replication and transcription in archaeans resembles that of eukarya, whereas their metabolism strongly resembles that of bacteria. DNA sequence analysis indicates that about half of the genes found in the kingdom Archaea are unique to this group. 3. Eukarya. This group includes all organisms whose cells contain an elaborate network of internal membranes, a membrane-bounded nucleus, and mitochondria. Their DNA is

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

present in the form of linear molecules organized into true chromosomes, and cell division takes place by means of mitosis (discussed in The Chromosomal Basis of Inheritance chapter). The eukaryotes include plants and animals as well as fungi and many single-celled organisms, such as amoebae and ciliated protozoa.

Genomes and Proteomes The totality of DNA in a cell, nucleus, or organelle is called its genome. When used with reference to a species of organism (for example, in phrases such as “the human genome”), the term genome is defined as the DNA present in a normal reproductive cell.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Modern methods for sequencing DNA (discussed in the DNA Replication and Sequencing chapter) are so rapid and efficient that the complete DNA sequence is now known for more than 1000 different species of organisms. These include the genomes of multiple representatives of all of the groups of organisms, including Neandertals (an extinct species closely related to Homo sapiens, sequenced from DNA extracted from fossil bones). TABLE 1.2 shows a small sample of sequenced genomes. Genome size is given in megabases (Mb), or millions of base pairs. The genome of the bacterium Haemophilus influenzae, like that of most bacteria, is very compact in that most of it codes for proteins. A high density of genes, relative to the amount of DNA, is also found in the budding yeast, the nematode worm, the fruit fly, and the diminutive flowering plant Arabidopsis thaliana. The human genome, by contrast, contains large amounts of noncoding DNA. Comparison with the nematode is illuminating. Whereas the human genome is about 30 times larger than that of the worm, the number

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

of genes is less than 2 times larger. This discrepancy reflects the fact that only about 1.5 percent of the human genome sequence codes for protein. (Approximately 27 percent of the human genome is present in genes, but much of the DNA sequence present in genes does not code for proteins.) TABLE 1.2 Comparison of genes and proteins Organism

Genome size,

Number of

Number of

Shared

Mba

genes

distinct

protein

(approximate)

(approximate)

proteins in

families

proteome b (approximate) Haemophilus

1.9

1700

1400

13

6000

4400

100

20,000

9500

} 3000

120c

16,000

8000

} 5000

2500

25,000

10,000

} 7000d

2900e

25,000

10,000

} 9900f

influenzae (causes bacterial meningitis) Saccharomyces cerevisiae (budding yeast) Caenorhabditis elegans (soil nematode)

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Drosophila melanogaster (fruit fly) Mus musculus (laboratory mouse) Homo sapiens (human being)

a Millions of base pairs.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

b Excludes “families” of proteins with similar sequences (and hence related functions). c Excludes 60 Mb of specialized DNA (“heterochromatin”) that has a very low content of

genes. d Based on similarity with sequences in messenger RNA (mRNA). e For convenience, this estimate is rounded to 3000 Mb elsewhere in this text. f Based on the observation that only about 1 percent of mouse genes lack a similar

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

gene in the human genome, and vice versa.

The complete set of proteins encoded in the genome is known as the proteome. In less complex genomes, such as the yeast, worm, and fruit fly, the number of proteins in an organism's proteome is approximately the same as the number of genes. However, as we shall see the chapter titled The Molecular Biology of Gene Expression, some genes encode two or more proteins through a process called alternative splicing in which segments of the original RNA transcript are joined together in a variety of combinations to produce different messenger RNAs. Alternative splicing is especially prevalent in the human genome. At least one third of human genes, and possibly as many as two thirds, undergo alternative splicing; among the genes that undergo alternative splicing, the number of different messenger RNAs per gene ranges from 2 to 7. Hence, with its seemingly limited repertoire of 25,000 genes, the human genome can create approximately 60,000 to 90,000 different mRNAs. The widespread use of alternative splicing to multiply the coding capacity of genes is one source of human genetic complexity. Most eukaryotic organisms contain families of related proteins that can be grouped according to similarities in their amino acid sequence. These families exist because the evolution of a new gene function is typically preceded by the duplication of an existing gene, followed by changes in nucleotide sequence in one of the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

copies that gives rise to the new function. The new function is usually similar to the previous one (for example, a change in the substrate specificity of a transporter protein), so that the new protein retains enough similarity in amino acid sequence to the original that their common ancestry can be recognized.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The molecular unity of life can be seen in the similarity of proteins in the proteome among diverse types of organisms. Such comparisons are shown in the right-hand column in Table 1.2. In this tabulation, each family of related proteins is counted only once, to estimate the number of proteins in the proteome that are “distinct” in the sense that their sequences are dissimilar. In yeast, worms, and flies, the number of distinct proteins is approximately 4400, 9500, and 8000, respectively. The brackets in Table 1.2 indicate the number of distinct proteins that share sequence similarity between species. From these comparisons, it appears that most multicellular animals share 5000 to 10,000 proteins that are similar in sequence and function. Approximately 3000 of these are shared with eukaryotes as distantly related as yeast, and approximately 1000 with prokaryotes as distantly related as bacteria. What these comparisons among proteomes imply is that biological systems are based on protein components numbering in the thousands. This is a challenging level of complexity to understand, but the challenge is much less intimidating than it appeared to be at an earlier time when human cells were thought to produce as many as 1 million different proteins.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

SUMMING UP ■ The processes of DNA replication, transcription, and translation, as well as the genetic code, are nearly universal properties of life. ■ Cellular organisms are divided into three domains—prokarya, eukarya, and archaea. ■ The genomes of organism vary greatly in total size, but less so in terms of the number of proteins encoded by them. ■ The proteome of an organism consists of families of related proteins, many of which are shared among organisms in different domains of life.

CHAPTER SUMMARY ■ Inherited traits are affected by genes. ■ Genes are composed of the chemical deoxyribonucleic acid (DNA).

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ DNA replicates to form copies of itself that are identical (except for rare mutations). ■ DNA contains a genetic code specifying which types of enzymes and other proteins are made in cells. ■ DNA occasionally mutates, and the mutant forms specify altered proteins that have reduced activity or stability. ■ A mutant enzyme is an “inborn error of metabolism” that blocks one step in a biochemical pathway for the metabolism of small molecules. ■ Traits are affected by environment as well as by genes. ■ The molecular unity of life is seen in comparisons of genomes and proteomes.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

REVIEW THE BASICS ■ What were the key experiments showing that DNA is the genetic material? ■ How did understanding the molecular structure of DNA give clues to its ability to replicate, to code for proteins, and to undergo mutation? ■ Why is the pairing of complementary bases a key feature of DNA replication? ■ What is the process of transcription, and in which ways does it differ from DNA replication? ■ What is the difference between transcription and translation? ■ Which three types of RNA participate in protein synthesis, and what is the role of each type of RNA? ■ What is the “genetic code,” and how is it relevant to the translation of a polypeptide chain from a molecule of messenger RNA? ■ What is an inborn error of metabolism? How did this concept serve as a bridge between genetics and biochemistry?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ How does the “central dogma” explain Garrod's discovery that nonfunctional enzymes result from mutant genes? ■ Explain why many mutant forms of phenylalanine hydroxylase have a simple amino acid replacement, yet the mutant polypeptide chains are absent or present in very small amounts. ■ What is a pleiotropic effect of a gene mutation? Give an example. ■ What are some of the major differences in cellular organization among prokarya, archaea, and eukarya?

GUIDE TO PROBLEM SOLVING

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

PROBLEM 1 In the human gene for the protein huntingtin (so named because it is associated with Huntington disease), the first 21 nucleotides in the amino acid—coding region are:

What is the sequence of a partner strand? ANSWER The base pairing between the strands is A with T and C with G, but it is equally important that the strands in a DNA duplex have opposite polarity. The complementary strand is therefore oriented with its 5′-end at the left. The base sequence of the partner strand is:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

PROBLEM 2 If the DNA duplex for huntingtin in Problem 1 were transcribed from left to right, deduce the base sequence of the RNA in this coding region. ANSWER To deduce the RNA sequence, we must apply three concepts. First, the base pairing is such so that A, T, C, and G in the DNA template strand are transcribed as U, A, G, and C, respectively, in the RNA. Second, the DNA template strand and RNA transcript have the opposite polarity. Third (and critically for this problem), the RNA strand is always transcribed in the 5′-to-3′ direction. Because we are told that transcription takes place from left to right, we can deduce that the transcribed strand is that given in Problem 1. The RNA transcript therefore has the base sequence

PROBLEM 3 Given the RNA sequence coding for part of human huntingtin deduced in Problem 2, what is the amino acid sequence in this part of huntingtin?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

ANSWER The polypeptide chain is translated in successive groups of three nucleotides (codons), starting at 5′ end of the coding sequence and moving in the 5′-to-3′ direction. The amino acid corresponding to each codon can be found in the genetic code table. The first seven amino acids in the polypeptide chain are:

PROBLEM 4 Suppose that a mutation in the human huntingtin gene occurs in the DNA sequence shown in Problem 1. In this mutation, the red G is replaced with a T. What is the nucleotide sequence of this region of the DNA duplex (both strands), and that of the messenger RNA, and what is the amino acid sequence of the mutant huntingtin?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

ANSWER The DNA, RNA, and polypeptide chain have the sequences as follows, where the differences from the nonmutant gene are in red. The mutation results in stop codon and premature termination of huntingtin synthesis.

ANALYSIS AND APPLICATIONS 1.1 Classify each of the following statements as true or false. (a) Each gene is responsible for only one visible trait.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

(b) Every trait is potentially affected by many genes. (c) The sequence of nucleotides in a gene specifies the sequence of amino acids in a protein encoded by the gene. (d) There is one-to-one correspondence between the set of codons in the genetic code and the set of amino acids encoded.

1.2 From their examination of the structure of DNA, what were Watson and Crick able to infer about the probable mechanisms of DNA replication, coding capability, and mutation? 1.3 What does it mean to say that each strand of a duplex DNA molecule has a polarity? What does it mean to say that the paired strands in a duplex molecule have opposite polarity?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1.4 What important observation about the S and R strains of Streptococcus pneumoniae prompted Avery, MacLeod, and McCarty to study this organism? 1.5 In the transformation experiments of Avery, MacLeod, and McCarty, what was the strongest evidence that the substance responsible for the transformation was DNA rather than protein? 1.6 A chemical called phenol (carbolic acid) destroys proteins but not nucleic acids, and strong alkalis such as sodium hydroxide destroy both proteins and nucleic acids. In the transformation experiments with Streptococcus pneumoniae, which result would be

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

expected if the S-strain extract had been treated with phenol? If it had been treated with a strong alkali? 1.7 Which feature of the physical organization of bacteriophage T2 made it suitable for use in the Hershey Chase experiments? 1.8 Like DNA, molecules of RNA contain large amounts of phosphorus. When Hershey and Chase grew their T2 phage in bacterial cells in the presence of radioactive phosphorus, the RNA must also have incorporated the labeled phosphorus, yet the experimental result was not compromised. Why not? 1.9 The DNA extracted from a bacteriophage contains 16 percent A, 16 percent T, 34 percent G, and 34 percent C. What can you conclude about the structure of this DNA molecule?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1.10 The DNA extracted from a bacteriophage consists of 20 percent A, 34 percent T, 35 percent G, and 11 percent C. What is unusual about this DNA? What can you conclude about its structure? 1.11 A region along a DNA strand that is transcribed contains no A. Which base will be missing in the corresponding region of the RNA? 1.12 If one strand of a DNA duplex has the sequence 5′ATCAG-3′, what is the sequence of the complementary strand? (Write the answer with the 5′ end at the left.) 1.13 Suppose that a double-stranded DNA molecule is separated into its constituent strands, and the strands are purified using a high-speed centrifuge. In

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

one of the strands, the base composition is 25 percent A, 18 percent T, 20 percent G, and 37 percent C. What is the base composition of the other strand? 1.14 Consider a region along one strand of a doublestranded DNA molecule that consists of tandem repeats of the trinucleotide 5′-GTA-3′, so that the sequence in this strand is 5′-GTAGTAGTAGT…-3′. What is the sequence in the other strand? (Write the answer with the 5′ end at the left.) 1.15 If the template strand of a DNA duplex has the sequence 5′-TCAG-3′, what is the sequence of the RNA transcript? (Write the answer with the 5′ end at the left.)

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1.16 If the nontemplate strand of a DNA duplex has the sequence 5′-ATCAG-3′, what is the sequence of the RNA transcript across this region? (Write the answer with the 5′ end at the left). 1.17 A duplex DNA molecule contains a random sequence of the four nucleotides with equal proportions of each. What is the average spacing between consecutive occurrences of the sequence 5′-ATGC3′? Between consecutive occurrences of the sequence 5′-TACGGC-3′? 1.18 An RNA molecule folds back upon itself to form a “hairpin” structure held together by a region of base pairing. One segment of the molecule in the paired region has the base sequence 5′-UAUCGUAU-3′. What is the base sequence with which this segment is paired?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.19 A synthetic mRNA molecule consists of the repeating base sequence

When this molecule is translated in vitro using ribosomes, transfer RNAs, and other necessary constituents from E. coli, the result is a polypeptide chain consisting of the repeating amino acid Lys–Lys–Lys–…. If you assume that the genetic code is a triplet code, what does this result imply about the codon for lysine (Lys)?

1.20 A synthetic mRNA molecule consisting of the repeating base sequence

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

is terminated by the addition, to the right-hand end, of a single nucleotide bearing A. When translated in vitro, the resulting polypeptide consists of a repeating sequence of phenylalanines terminated by a single leucine. What does this result imply about the codon for leucine?

1.21 With in vitro translation of an RNA into a polypeptide chain, the translation can begin anywhere along the RNA molecule. A synthetic RNA molecule has the sequence

How many reading frames are possible if this molecule is translated in vitro? How many reading frames are possible if this molecule is translated in vivo, in which translation starts with the codon AUG?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.22 You have sequenced both strands of a doublestranded DNA molecule. To evaluate the potential of this molecule for coding amino acids, you conceptually transcribe it into RNA and then conceptually translate the RNA into a polypeptide chain. How many reading frames would you have to examine? 1.23 A synthetic mRNA molecule consists of the repeating base sequence

When this molecule is translated in vitro, the result is a polypeptide chain consisting of the alternating amino acids Thr–His–Thr–His–Thr–His–…. Why do the amino acids alternate? What does this result imply about the codons for threonine (Thr) and histidine (His)?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1.24 A synthetic mRNA molecule consists of the repeating base sequence

When this molecule is translated in vitro, the result is a mixture of three different polypeptide chains. One consists of repeating isoleucines (Ile – Ile – Ile – Ile –…), another of repeating serines (Ser – Ser – Ser – Ser –…), and the third of repeating histidines (His – His – His –His –…). What does this result imply about the manner in which an mRNA is translated?

1.25 How is it possible for a gene with a mutation in the coding region to encode a polypeptide with the same amino acid sequence as the nonmutant gene?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

1.26 A polymer has a random sequence consisting of 75 percent G and 25 percent U. Among the amino acids in the polypeptide chains resulting from in vitro translation, what is the expected frequency of Trp? Of Val? Of Phe?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1.27 Results of complementation tests of the six independent recessive mutations a–f are shown in the accompanying matrix, where + indicates complementation and − indicates lack of complementation. Classify the mutations into complementation groups, and write the names of the mutations in each complementation group in the blanks provided below. (Some of the blanks may remain empty.)

Mutations in complementation group 1: ___ ___ ___ ___ ___ Mutations in complementation group 2: ___ ___ ___ ___ ___ Mutations in complementation group 3: ___ ___ ___ ___ ___ Mutations in complementation group 4: ___ ___ ___ ___ ___

1.28 A mutant screen is carried out in Neurospora crassa for mutants that are unable to synthesize an amino acid, that we will symbolize as G. A number of mutants are isolated and classified into four groups according to their ability to grow (+) or not grow (−) in minimal medium supplemented with possible

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

precursors D, E, and F. The data are shown in the accompanying table. D

E

F

G

Class 1

+

+

+

+

Class 2

−

−

−

+

Class 3

−

−

+

+

Class 4

+

−

+

+

Complete the diagram shown below. Each arrow indicates one or more biochemical reactions. Within each circle, write the class of mutants (1–4) whose products contribute to the reactions symbolized by the arrow; in the squares, write the name of the amino acid or precursor (D–G) at that position in the pathway.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1.29 The coding sequence in the messenger RNA for amino acids 1 through 10 of human phenylalanine hydroxylase is:

(a) What are the first 10 amino acids? (b) Which sequence would result from a mutant RNA in which the red A was changed to G? (c) Which sequence would result from a mutant RNA in which the red C was changed to G?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

(d) Which sequence would result from a mutant RNA in which the red U was changed to C? (e) Which sequence would result from a mutant RNA in which the red G was changed to U?

1.30 A “frameshift” mutation is a mutation in which some number of base pairs, other than a multiple of three, is inserted into or deleted from a coding region of DNA. The result is that, at the point of the frameshift mutation, the reading frame of protein translation is shifted with respect to the nonmutant gene. To see the consequence of a frameshift mutation, consider that the coding sequence in the messenger RNA for the first 10 amino acids in human beta hemoglobin (part of the oxygen-carrying protein in the blood) is

(a) What is the amino acid sequence in this part of the polypeptide chain?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(b) What would be the consequence of a frameshift mutation resulting in an RNA missing the red U? (c) What would be the consequence of a frameshift mutation resulting in an RNA with a G inserted immediately in front of the red U?

1.31 A news headline reads “Gene for Schizophrenia Identified.” Does this necessarily mean that schizophrenia in an individual can be diagnosed by genetic analysis alone? Explain your answer. 1.32 Using the data in Table 1.2, construct a graph, plotting genome size for the different organisms on

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

the x-axis and number of genes on the y-axis. Now, consider the case of the Mexican axolotl (Ambystoma mexicana), with an estimated genome size of 32 billion base pairs. Based on your graph, can you predict the number of genes that are present in this genome? Explain your answer.

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 With regard to the wild-type and mutant RNA molecules described in Problem 1.30, deduce the base sequence in both strands of the corresponding doublestranded DNA for: (a) The wild-type sequence (b) The single-base deletion (c) The single-base insertion

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

CHALLENGE PROBLEM 2 You are carrying out Beadle Tatum type experiments to analyze a metabolic pathway in Neurospora. You know that the precursor in the pathway is a molecule symbolized as P and that the product is a vitamin symbolized as Z. You are sure that the pathway from P to Z is linear and that the molecules W, X, and Y are intermediates. However, there may be other intermediates not yet identified. You obtain 10 independent mutations that cannot grow on minimal medium supplemented with P, but can grow on minimal medium supplemented with Z. The 10 mutants fall into four classes that can grow (+) or cannot grow (−) on minimal medium supplemented with the nutrients W, X, or Y. The data are shown in the accompanying table.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

P

W

X

Y

Z

Class I (mutant 5)

−

−

−

−

+

Class II (mutants 1, 3, 4, 6, 7)

−

+

+

+

+

Class III (mutant 2)

−

+

−

+

+

Class IV (mutants 8, 9, 10)

−

−

−

+

+

Draw a linear metabolic pathway with P on the left and Z on the right, in which each of the intermediates W, X, and Y is shown in the order in which it occurs in the metabolic pathway in the synthesis of Z from P.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

CHALLENGE PROBLEM 3 The 10 mutants in Challenge Problem 2 were tested for complementation in all pairwise combinations using heterokaryons. The results are shown in the matrix, in which + indicates the ability of the heterokaryon to grow in minimal medium and − indicates inability to grow in minimal medium.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Mutant 1 2 3 4 5

2

3

4

5

6

7

8

9

10

1

1

2

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

2

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

6 7 8 9

2

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Assume that each complementation group defines a different gene, and assume further that each gene encodes an enzyme that catalyzes a single step in the metabolic pathway, which converts one molecule of substrate into one molecule of product. a. (a) Redraw the metabolic pathway deduced from Challenge Problem 2. Use a right arrow to indicate each enzymatic step in the pathway, and label each arrow with the mutant number 1–10 that blocks the enzymatic step. In some cases, you will not be able to specify the order in which the enzymes occur in the pathway, so you may write them in any order you wish. b. (b) In the metabolic pathway that you have deduced from the data, how many unknown intermediates are there between the precursor P and the vitamin Z?

FOR FURTHER READING

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

The 1000 Genomes Consortium. (2015). A global reference for human genetic variation. Nature, 526(7571(68–74 http://doi.org/10.1038/nature15393 An overview of the 1000 Genomes Project. Avery, O. T. (1944). Studies on the chemical nature of the substance inducing transformation of pneumococcal types: Induction of transformation by a deoxyribonucleic acid fraction isolated from pneumococcus type III. Journal of Experimental Medicine, 79(2), 137–158. http://doi.org/10.1084/jem.79.2.137 Considered by most to be the definitive evidence that DNA is the genetic material.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Beadle, G. W., & Tatum, E. L. (1941). Genetic control of biochemical reactions in Neurospora. Proceedings of the National Academy of Sciences of the United States of America, 2(11), 499–506. http://www.ncbi.nlm.nih.gov/pubmed/16588492 The origin of the one gene–one enzyme hypothesis. Crick, F. (1970). Central dogma of molecular biology. Nature, 227(5258), 561–563. http://doi.org/10.1038/227561a0

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Crick's clearest statement of the central dogma, in terms of information flow from DNA to protein but not the reverse. Griffith, F. (1928). The significance of pneumococcal types. Journal of Hygiene, 27(2), 113. http://doi.org/10.1017/S0022172400031879 The first demonstration that there is a “transforming principle.” Hershey, A. D., & Chase, M. (1952). Independent functions of viral protein and nucleic acid in growth of bacteriophage. Journal of General Physiology, 36(1), 39–56. http://doi.org/10.1085/jgp.36.1.39

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Further confirmation that DNA, as opposed to protein, is the genetic material. Watson, J. D., & Crick, F. H. (1953). Molecular structure of nucleic acids: A structure for deoxyribose nucleic acid. Nature, 171(4356), 737– 738. http://www.ncbi.nlm.nih.gov/pubmed/13054692 The original publication describing the double-helical structure of DNA. Watson, J. D., & Crick, F. H. (1953). Genetical implications of the structure of deoxyribonucleic acid. Nature. http://doi.org/10.1038/171964b0

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

A less well-known but nevertheless very important paper in which Watson and Crick first suggest a basis for the genetic code.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:29.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

© Science Source.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

CHAPTER 2 DNA Structure and Genetic Variation CHAPTER OUTLINE 2.1 Genetic Differences Among Individuals 2.2 The Terminology of Genetic Analysis 2.3 The Molecular Structure of DNA 2.4 The Separation and Identification of Genomic DNA Fragments

2.5 Amplification of Specific DNA for Detection and Purification

2.6 Types of DNA Markers Present in Genomic DNA 2.7 Applications of DNA Markers

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

ROOTS OF DISCOVERY: The Double Helix James D. Watson and Francis H. C. Crick (1953) A Structure for Deoxyribose Nucleic Acid THE CUTTING EDGE: High-Throughput SNP Genotyping

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

LEARNING OBJECTIVES & SCIENCE COMPETENCIES Application of the principles of DNA structure and genetic variation examined in this chapter will enable you to solve the following types of problems: ■ Explain the structure of DNA and how that structure facilitates its manipulation. ■ Produce a restriction map of a DNA fragment based on gel patterns of fragments produced by cleaving it with multiple restriction enzymes. ■ Use data from analysis of DNA markers to identify individuals and relatedness.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ For a given sequence of DNA, select the sequences of primer oligonucleotides that would allow any specific fragment of the molecule to be amplified in the polymerase chain reaction. rior to the mid-1970s, classical and molecular genetics, while addressing the same questions, were often treated as separate disciplines. Since then, studies in genetics have undergone a revolution based on the use of increasingly sophisticated ways to isolate and identify specific fragments of DNA, which have substantially merged the two approaches. The culmination of these developments was large-scale genomic sequencing—the ability to determine the correct sequence of the base pairs that make up the DNA in an entire genome and to identify the sequences associated with genes. Because many of

P

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

the laboratory organisms used in genetics experiments have relatively small genomes, these sequences were completed first. The techniques used to sequence these simpler genomes were then scaled up to sequence much larger genomes, including the human genome. This has greatly expanded our ability to investigate

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

even the most complex of traits at the molecular, cellular, and organismal levels.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.1 Genetic Differences Among Individuals The human genome in a reproductive cell consists of approximately 3 billion base pairs organized into 23 distinct chromosomes (each chromosome contains a single molecule of duplex DNA). A typical chromosome can contain several hundred to several thousand genes, arranged in linear order along the DNA molecule present in the chromosome. The sequences that make up the protein-coding

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

part of these genes actually account for only about 1.3 percent of the entire genome. The other 98.7 percent of the sequences do not code for proteins. Some encode RNAs that are not mRNAs, but rather are molecules that contribute to a wide variety of cellular functions. Others are noncoding sequences, with many of these being relatively short sequences that are found in hundreds of thousands of copies scattered throughout the genome. Still other noncoding sequences are decayed remnants of genes called pseudogenes. And still others are noncoding sequences whose functions are the subject of much current investigation. As might be expected, identifying the protein-coding genes against the large background of noncoding DNA in the human genome is a challenge in itself. Geneticists often speak of the nucleotide sequence of “the” human genome because corresponding DNA sequences from any two individuals are identical at approximately 99.9 percent of their nucleotide sites. These shared sequences are our evolutionary legacy: They contain the genetic information that makes us human beings. In reality, however, there are many different human genomes. Geneticists have the most interest in the 0.1 percent of the human DNA sequence—3 million base pairs—that differs from one genome to the next. Most of this variation is “normal,” but these differences also include the mutations that are responsible for genetic diseases such as phenylketonuria (PKU) and other

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

inborn errors of metabolism, as well as the mutations that increase individuals' risk of developing more complex diseases such as heart disease, breast cancer, and diabetes. Fortunately, only a small proportion of all differences in DNA sequence are associated with disease. Some of the others are associated with inherited differences in height, weight, hair color, eye color, facial features, and other traits. Most of the genetic differences between people are completely harmless. Many have no detectable effects on appearance or health. Such differences can be studied only through direct examination of the DNA itself. These differences are nevertheless important, because they serve as genetic markers.

DNA Markers as Landmarks in

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Chromosomes In genetics, a genetic marker is any difference in DNA, no matter how it is detected, whose pattern of transmission from generation to generation can be tracked. Each individual who carries the marker also carries a length of chromosome on either side of it, so that the marker marks a particular region of the genome. Any difference in DNA sequence between two individuals can serve as a genetic marker. Although genetic markers are often harmless in themselves, they allow the positions of disease genes to be located along the chromosomes and their DNA to be isolated, identified, and studied. Genetic markers that are detected by direct analysis of the DNA are often called DNA markers. DNA markers are important in genetics because they serve as landmarks in long DNA molecules, such as those found in chromosomes, which allow researchers to track genetic differences among individuals. In this sense, they are

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

like signposts along a highway. Using DNA markers as landmarks, geneticists can identify the positions of normal genes, mutant genes, breaks in chromosomes, and other features important in genetic analysis. A hypothetical example is given in FIGURE 2.1, involving the PAH gene encoding phenylalanine hydroxylase. In this figure, we see examples of two copies of the segment of chromosome 12, one of which codes for the normal protein and the other of which codes

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

for a protein associated with PKU. These two segments also differ with respect to a noncoding nucleotide. This nucleotide could be a genetic marker, in that while the G residue associated with the PKU variant is not the difference that causes PKU, its presence on a particular chromosome could be used as a marker to indicate that variant's presence.

FIGURE 2.1 This schematic shows two DNA molecules that contain both PAH variants and a DNA marker—in this case, an A or a G at the indicated position. In this hypothetical case, the A allele is associated with the disease-associated variant and, therefore, could be used as an indicator of the latter's presence in an individual's genome.

Of course, the problem we face is how to start from the total DNA of an individual and isolate a particular gene of interest, so that we can identify genetic differences between individuals. In this chapter, we will examine some of the principal ways in which DNA is manipulated to achieve this feat, whether or not these differences result in observable differences. An overview of the steps involved

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

is shown in FIGURE 2.2. Use of these methods broadens the scope of genetics, making it possible to carry out genetic analysis in any organism. As a consequence, detailed genetic analysis is no longer restricted to human beings, domesticated animals, cultivated plants, and the relatively small number of model organisms favorable for genetic studies. Direct study of DNA eliminates the need for prior identification of genetic differences between individuals; it even eliminates the need for controlled crosses. The methods of molecular analysis discussed in this chapter have

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

transformed genetics and are the principal techniques used in almost every modern genetics laboratory; furthermore, having a basic understanding of them makes it possible to appreciate the overall unity of classical and molecular genetics.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.2 DNA markers serve as landmarks that identify physical positions along a DNA molecule, such as DNA from a chromosome. A DNA marker can also be used to identify bacterial cells into which a particular fragment of DNA has been introduced. The procedure of DNA cloning is not quite as simple as indicated here; it is discussed further in the Manipulating Genes and Genomes chapter.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

SUMMING UP ■ Every human genome is unique, but only a small fraction of the genome differs from individual to individual. ■ Most of the variation that does exist has no obvious physical effect. Variants associated with hereditary diseases are rare. ■ Protein-coding genes represent only a small fraction of the DNA in a human genome.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Studying DNA markers greatly extends the power of genetic analysis.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.2 The Terminology of Genetic Analysis To discuss genetic analysis at any level, we must first introduce some key terms that provide the essential vocabulary of genetics. These terms can be understood with reference to FIGURE 2.3. In the Genes, Genomes, and Genetic Analysis chapter, we defined a gene as an element of heredity, transmitted from parents to offspring in reproduction, that influences one or more hereditary

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

traits. Chemically, a gene is a sequence of nucleotides along a DNA molecule. In a population of organisms, not all copies of a gene may have exactly the same nucleotide sequence. For example, whereas one form of a gene may have the codon GCA at a certain position, another form of the same gene may have the codon GCG. Both codons specify alanine. Hence, the two forms of the gene encode the same sequence of amino acids, yet differ in DNA sequence. The alternative forms of a gene are called alleles of the gene. Different alleles may also code for different amino acid sequences, sometimes with drastic effects. Recall the example of the PAH gene for phenyalanine hydroxylase in the Genes, Genomes, and Genetic Analysis chapter, in which a change in codon 408 from CGG (arginine) to TGG (tryptophan) results in an inactive enzyme that becomes expressed as the inborn error of metabolism phenylketonuria.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE 2.3 Key concepts and terms used in modern genetics. Note that a single gene can have any number of alleles in the population as a whole, but no more than two alleles can be present in any one individual.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Within a cell, genes are arranged in linear order along microscopic thread-like bodies called chromosomes, which we examine in detail in the chapters titled The Chromosomal Basis of Inheritance and Human Karyotypes and Chromosome Behavior. Each human reproductive cell contains one complete set of 23 chromosomes containing 3 × 109 base pairs of DNA. A typical chromosome contains several hundred to several thousand genes. In humans, the average is approximately 1000 genes per chromosome. Each chromosome contains a single molecule of duplex DNA along its length, complexed with proteins and very tightly coiled. The DNA in the average human chromosome, when fully extended, has relative dimensions comparable to those of a wet spaghetti noodle 25 miles long; when the DNA is coiled in the form of a chromosome, its physical compaction is comparable to that of the same noodle coiled and packed into an 18-foot canoe. The physical position of a gene along a chromosome is called the locus of the gene. In most higher organisms, including human

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

beings, each cell other than a sperm or egg contains two copies of each type of chromosome—one inherited from the mother and one inherited from the father. Each member of such a pair of chromosomes is said to be homologous to the other. (The chromosomes that determine sex are an important exception, which we will ignore for now.) At any locus, therefore, each individual carries two alleles, because one allele is present at a corresponding position in each of the homologous maternal and paternal chromosomes (Figure 2.3). The genetic constitution of an individual is called its genotype. For a particular gene, if the two alleles at the locus in an individual are indistinguishable from each other, then the genotype of the individual is said to be homozygous for the allele that is present. If the two alleles at the locus are different from each other, then the genotype of the individual is said to be heterozygous for the alleles that are present. Typographically, genes are indicated in italics, and alleles are typically distinguished by uppercase or lowercase letters (A versus a), subscripts (A1 versus A2), superscripts (a+ versus a−), or sometimes just + and −. Using

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

these symbols, homozygous genes would be portrayed by any of these formulas: AA, aa, A1A1, A2A2, a+a+, a−a−, +/+, or −/−. As in the last two examples, the slash is sometimes used to separate alleles present in homologous chromosomes to avoid ambiguity. Heterozygous genes would be portrayed by any of the formulas Aa, A1A2, a+a−, or +/−. In Figure 2.3, the genotype Bb is heterozygous because the B and b alleles are distinguishable (which is why they are assigned different symbols), whereas the genotype CC is homozygous. These genotypes could also be written as B/b and C/C, respectively. Whereas the alleles that are present in an individual constitute its genotype, the physical or biochemical expression of the genotype

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

is called the phenotype. To put it as simply as possible, the distinction is that the genotype of an individual is what is on the inside (the alleles in the DNA), whereas the phenotype is what is on the outside (the observable traits, including biochemical traits, behavioral traits, and so forth). The distinction between genotype and phenotype is critically important because there usually is not a one-to-one correspondence between genes and traits. Most complex traits—such as hair color, skin color, height, weight, behavior, life span, and susceptibility to disease—are influenced by

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

many genes. Most traits are also influenced more or less strongly by environment. Thus, depending on the environment, the same genotype can result in different phenotypes. Compare, for example, two people with a genetic risk for lung cancer: If one smokes and the other does not, the smoker is much more likely to develop the disease. Environmental effects also imply that the same phenotype can result from more than one genotype. Smoking again provides an example, because most smokers who are not genetically at risk can also develop lung cancer. Harmful mutations represent only a small fraction of the total allelic variation in a species, and most such alleles are very uncommon. In contrast, in many cases, different alleles at a locus can have no evident effects on phenotypes; often multiple alleles of such loci are found at relatively high frequencies (greater than 5 percent). Such cases are called polymorphisms; the term polymorphism literally means “multiple forms.” At the sequence level, these variants are referred to as DNA polymorphisms.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

SUMMING UP ■ The genome of a typical human cell contains 23 pairs of homologous chromosomes, each of which carries an average of 1000 genes. ■ Variants of particular genes are known as alleles. ■ Individuals with two identical alleles of a particular gene are homozygotes; those with two different alleles are heterozygotes. ■ The combination of alleles carried by an individual is the genotype. ■ The physical manifestation of the genotype is the phenotype.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Many loci in the human genome are polymorphic.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.3 The Molecular Structure of DNA Modern experimental methods for the manipulation and analysis of DNA grew out of a detailed understanding of its molecular structure and replication. Therefore, to understand these methods, one needs to know something about the molecular structure of DNA. As we saw in the Genes, Genomes, and Genetic Analysis chapter, DNA is a helix consisting of two paired, complementary strands, each composed of an ordered string of nucleotides, with each nucleotide bearing one of the bases A (adenine), T (thymine), G (guanine), or C (cytosine). Watson–Crick base pairing between A and T and between G and C in the complementary strands holds the strands together. The complementary strands also hold the key to replication, because each strand can serve as a template for the synthesis of a new complementary strand. We will now take a closer look at DNA structure and at the key features of its replication.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Polynucleotide Chains In terms of biochemistry, a DNA strand is a polymer—a large molecule built from repeating units. The units in DNA are composed of 2′-deoxyribose (a five-carbon sugar), phosphoric acid, and the four nitrogen-containing bases denoted A, T, G, and C. The chemical structures of the bases are shown in FIGURE 2.4. Note that two of the bases have a double-ring structure; these are called purines. The other two bases have a single-ring structure; these are called pyrimidines.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE 2.4 Chemical structures of the four nitrogen-containing bases in DNA: adenine, thymine, guanine, and cytosine. The nitrogen atom linked to the deoxyribose sugar is indicated. The atoms shown in red participate in hydrogen bonding between the DNA base pairs.

■ The purine bases are adenine (A) and guanine (G).

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The pyrimidine bases are thymine (T) and cytosine (C). In DNA, each base is chemically linked to one molecule of the sugar deoxyribose, forming a compound called a nucleoside. When a phosphate group is also attached to the sugar, the nucleoside becomes a nucleotide (FIGURE 2.5). Thus, a nucleotide is a nucleoside plus a phosphate. In the conventional numbering of the carbon atoms in the sugar in Figure 2.3, the carbon atom to which the base is attached is the 1′ carbon. (The atoms in the sugar are given primed numbers to distinguish them from atoms in the bases.) The nomenclature of the nucleoside and nucleotide derivatives of the DNA bases is summarized in TABLE 2.1. Most of these terms are not needed in this text; they are included because they are likely to be encountered in further reading.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE 2.5 A typical nucleotide, showing the three major

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

components (phosphate, sugar, and base) that are the difference between DNA and RNA. A nucleoside consists of the sugar and base only. Nucleotides are monophosphates (with one phosphate group). Nucleoside diphosphates contain two phosphate groups, and nucleoside triphosphates contain three.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

TABLE 2.1 DNA nomenclature Base

Nucleoside

Nucleotide

Adenine (A)

Deoxyadenosine

Deoxyadenosine-5′  monophosphate (dAMP)  diphosphate (dADP)  triphosphate (dATP)

Guanine (G)

Deoxyguanosine

Deoxyguanosine-5′  monophosphate (dGMP)  diphosphate (dGDP)  triphosphate (dGTP)

Thymine (T)

Deoxythymidine

Deoxythymidine-5′  monophosphate (dTMP)  diphosphate (dTDP)  triphosphate (dTTP)

Cytosine (C)

Deoxycytidine

Deoxycytidine-5′  monophosphate (dCMP)  diphosphate (dCDP)

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

 triphosphate (dCTP)

In nucleic acids, such as DNA and RNA, the nucleotides join together to form a polynucleotide chain, in which the phosphate attached to the 5′ carbon of one sugar is linked to the hydroxyl group attached to the 3′ carbon of the growing chain (FIGURE 2.6). The chemical bonds by which the sugar components of adjacent nucleotides are linked through the phosphate groups are called phosphodiester bonds. The 3′–5′–3′–5′ orientation of these linkages continues throughout the chain, which typically consists of millions of nucleotides. Note that the terminal groups of each

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

polynucleotide chain are a 5′-phosphate (5′-P) group at one end

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

and a 3′-hydroxyl (3′-OH) group at the other end. The asymmetry of the ends of a DNA strand implies that each strand has a polarity determined by which end bears the 5′ phosphate and which end bears the 3′ hydroxyl.

FIGURE 2.6 Three nucleotides at the 5′ end of a single polynucleotide strand. (A) The chemical structure of the sugar– phosphate linkages, showing the 5′-to-3′ orientation of the strand (the red numbers are those assigned to the carbon atoms). (B) A common schematic way to depict a polynucleotide strand.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

A few years before Watson and Crick proposed their essentially correct three-dimensional structure of DNA as a double helix, Erwin Chargaff developed a chemical technique to measure the amount of each base present in DNA. As we describe this technique, we denote the molar concentration of any base by the symbol for the base enclosed in square brackets; for example, [A] denotes the molar concentration of adenine. Chargaff used his technique to measure the [A], [T], [G], and [C] content of the DNA from a variety of sources. He found that the base composition of the DNA, defined as the percent G + C, differs among species but is constant in all cells of an organism and within a species. Data on the base composition of DNA from a variety of organisms are given in TABLE 2.2.

TABLE 2.2 Base composition of DNA from different organisms Base (and percentage of total bases) Organism

Adenine

Thymine

Guanine

Cytosine

Base composition (percent G + C)

Bacteriophage T7

26.0

26.0

24.0

24.0

48.0

36.9

36.3

14.0

12.8

26.8

30.2

29.5

21.6

18.7

40.3

Escherichia coli

24.7

23.6

26.0

25.7

51.7

Sarcina lutea

13.4

12.4

37.1

37.1

74.2

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Bacteria Clostridium perfringens Streptococcus pneumoniae

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Fungi Saccharomyces

31.7

32.6

18.3

17.4

35.7

23.0

22.3

27.1

27.6

54.7

Wheat

27.3

27.2

22.7

22.8*

45.5

Maize

26.8

27.2

22.8

23.2*

46.0

30.8

29.4

19.6

20.2

39.8

Pig

29.4

29.6

20.5

20.5

41.0

Salmon

29.7

29.1

20.8

20.4

41.2

Human being

29.8

31.8

20.2

18.2

38.4

cerevisiae Neurospora crassa Higher plants

Animals Drosophila melanogaster

* Includes one-fourth 5-methylcytosine, a modified form of cytosine found in most plants

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

more complex than algae and in many animals.

Chargaff also observed certain regular relationships among the molar concentrations of the different bases. These relationships are now called Chargaff's rules: ■ The amount of adenine equals that of thymine: [A] = [T]. ■ The amount of guanine equals that of cytosine: [G] = [C]. ■ The amount of the purine bases equals that of the pyrimidine bases: [A] + [G] = [T] + [C].

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Although the chemical basis of these observations was not known at the time, one of the appealing features of the Watson–Crick structure of paired complementary strands was that it explained Chargaff's rules. Because A is always paired with T in doublestranded DNA, it must follow that [A] = [T]. Similarly, because G is paired with C, we know that [G] = [C]. The third rule follows by addition of the other two: [A] + [T] = [G] + [C]. In the next section, we examine the molecular basis of base pairing in more detail.

The Double Helix

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In addition to Chargaff's rules and the basic chemistry of nucleotides, Watson and Crick drew on structural observations made by Rosalind Franklin and Maurice Wilkins. These scientists used x-ray crystallography to examine the three-dimensional structure of DNA. In this procedure, crystals of large molecules are exposed to beams of x-rays. The shape of the crystals causes the x-rays to scatter, or diffract, in a manner that is determined by the structure of the crystal. The pattern of diffraction is then recorded on photographic film, and the crystallographer can use it to make inferences about the crystal's structure. Perhaps the most famous such image ever made was Franklin's Photo 51 (FIGURE 2.7). From it, crystallographers could make the following inferences: ■ The crystal has a helical structure. ■ The diameter of the helix is 20 angstroms (Å). ■ The helix has two levels of repetitive structure, or periodicities, along its length. One of these occurs every 3.4 Å, and the other occurs every 34 Å.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE 2.7 Rosalind Franklin's “Photo 51,” the x-ray diffraction pattern obtained from crystallized DNA. The cross-like nature of the image indicates the helical structure of DNA. © Science Source.

Watson and Crick used all of these data when building their model of the double helix.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Base Pairing and Base Stacking In the three-dimensional structure of the DNA molecule proposed in 1953 by Watson and Crick, the molecule consists of two polynucleotide chains twisted around each other to form a doublestranded helix in which adenine and thymine, and guanine and cytosine, are paired in opposite strands (FIGURE 2.8). In the standard structure, which is called the B form of DNA, each chain makes one complete turn every 34 Å. The helix is right-handed, which means that as one looks down the barrel, each chain follows a clockwise path as it progresses. The bases are spaced at 3.4 Å,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

so there are 10 bases per helical turn in each strand and 10 base

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

pairs per turn of the double helix. These correspond to the two periodicities inferred from the x-ray crystallographic data.

FIGURE 2.8 Two representations of DNA, illustrating the threedimensional structure of the double helix. (A) In a ribbon diagram, the sugar–phosphate backbones are depicted as bands, with horizontal lines used to represent the base pairs. (B) A computer model of the B form of a DNA molecule. The stick figures are the sugar–phosphate chains winding around outside the stacked base

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

pairs, forming a major groove and a minor groove. The color coding for the base pairs is as follows: A, red or pink; T, dark green or light green; G, dark brown or beige; C, dark blue or light blue. The bases depicted in dark colors are those attached to the blue sugar–phosphate backbone; the bases depicted in light colors are attached to the beige backbone. (B) Courtesy of Antony M. Dean, University of Minnesota.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The strands feature base pairing, in which each base is paired with a complementary base in the other strand by hydrogen bonds. (A hydrogen bond is a weak bond in which two participating atoms share a hydrogen atom between them.) The hydrogen bonds provide one type of force holding the strands together. In Watson–Crick base pairing, adenine (A) pairs with thymine (T), and guanine (G) pairs with cytosine (C). The hydrogen bonds that form in the adenine–thymine base pair and in the guanine–cytosine pair are illustrated in FIGURE 2.9. Note that an A–T pair (Figure 2.9A and B) has two hydrogen bonds and that a G–C pair (Figure 2.9C and D) has three hydrogen bonds. This means that the hydrogen bonding between G and C is stronger in the sense that it requires more energy to break; for example, the amount of heat required to separate the paired strands in a DNA duplex increases with the percentage of G + C. Because nothing restricts the sequence of bases in a single strand, any sequence could be present along one strand. This explains Chargaff's observation that DNA from different organisms may differ in base composition. However, because the strands in duplex DNA are complementary, Chargaff's rules of [A] = [T] and [G] = [C] are true whatever the base composition.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.





Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.9 Normal base pairs in DNA. On the left, the hydrogen bonds (dotted lines) with the joined atoms are shown in red. (A and B) A–T base pairing. (C and D) G–C base pairing. In the spacefilling models (B and D), the colors are as follows: C, gray; N, blue; O, red; and H (shown in the bases only), white. Each hydrogen bond is depicted as a white disk squeezed between the atoms sharing the hydrogen. The stick figures on the outside represent the backbones winding around the stacked base pairs. Source: (B, D) Space-filling models courtesy of Antony M. Dean, University of Minnesota.

In the B form of DNA, the paired bases are stacked on top of one another like pennies in a roll. The upper and lower faces of each nitrogenous base are relatively flat and nonpolar (uncharged). These surfaces are said to be hydrophobic because they bind

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

poorly to water molecules, which are very polar. (The polarity refers to the asymmetrical distribution of charge across the Vshaped water molecule; the oxygen atom at the base of the V tends to be quite negative, whereas the hydrogen atoms at the tips are quite positive). Owing to their repulsion of water molecules, the paired nitrogenous bases tend to stack on top of one another in such a way as to exclude the maximum amount of water from the interior of the double helix. This feature of double-stranded DNA is known as base stacking. A double-stranded DNA molecule

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

therefore has a hydrophobic core composed of stacked bases, and it is the energy of base stacking that provides double-stranded DNA with much of its chemical stability (base pairing, by itself, would be insufficient to stabilize the molecule under physiological conditions). When discussing a DNA molecule, molecular biologists frequently refer to the individual strands as single strands or as singlestranded DNA; they refer to the double helix as double-stranded DNA or a duplex molecule. The two grooves spiraling along outside of the double helix are not symmetrical; one groove, called the major groove, is larger than the other, which is called the minor groove. Proteins that interact with double-stranded DNA often have regions that make contact with the base pairs by fitting into the major groove, into the minor groove, or into both grooves (Figure 2.8B).

Antiparallel Strands Each backbone in a double helix consists of deoxyribose sugars alternating with phosphate groups that link the 3′ carbon atom of one sugar to the 5′ carbon of the next in line (Figure 2.5). The two polynucleotide strands of the double helix have opposite polarity, in the sense that the 5′ end of one strand is paired with the 3′ end of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

the other strand. Strands with such an arrangement are said to be antiparallel. One implication of the presence of antiparallel strands in duplex DNA is that in each pair of bases, one base is attached to a sugar that lies above the plane of pairing, and the other base is attached to a sugar that lies below the plane of pairing. Another

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

implication is that each terminus of the double helix possesses one 5′-P group (on one strand) and one 3′-OH group (on the other strand), as shown in FIGURE 2.10.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.10 A segment of a DNA molecule, showing the antiparallel orientation of the complementary strands. The shaded blue arrows indicate the 5′-to-3′ direction of each strand. The phosphates (P) join the 3′ carbon atom of one deoxyribose to the 5′ carbon atom of the adjacent deoxyribose.

The diagram of the DNA duplex in Figure 2.8 is static and, therefore, somewhat misleading. DNA is actually a dynamic molecule, constantly in motion. In some regions, the strands can separate briefly and then come together again in the same conformation or in a different one. The right-handed double helix in

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Figure 2.8 is the standard B form, but depending on conditions, DNA can actually form more than 20 slightly different variants of a right-handed helix, and some regions can even form helices in which the strands twist to the left (called the Z form of DNA). If complementary stretches of nucleotides appear in the same strand, then a single strand, separated from its partner, can fold back upon itself like a hairpin. Even triple helices consisting of three strands can form in regions of DNA that contain suitable base sequences.

DNA Structure as Related to Function

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In the structure of the DNA molecule, we can see how the four essential requirements of a genetic material are met. 1. Any genetic material must be able to be replicated accurately, so that the information it contains will be precisely replicated and inherited by daughter cells. The basis for exact duplication of a DNA molecule is the pairing of A with T and of G with C in the two polynucleotide chains. Unwinding and separation of the strands, with each free strand being copied, results in the formation of two identical double helices. 2. Genetic material must carry encoded information. Quite clearly, the order of the bases in DNA provides the basis for such a code—specifically, a genetic code in which groups of three bases specify amino acids. Because the four bases in a DNA molecule can be arranged in any sequence, and because the sequence can vary from one part of the molecule to another and from organism to organism, DNA can contain a great many unique regions, each of which can be a distinct gene. 3. Genetic material must have the capacity to direct the organization and metabolic activities of the cell. As we saw

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

in the Genes, Genomes, and Genetic Analysis chapter, genes can direct the synthesis of a protein molecule—a polymer composed of repeating units of amino acids. The sequence of amino acids in the protein determines its chemical and physical properties. A gene is expressed when its protein product is synthesized, and one requirement of the genetic material is that it direct the sequence in which amino acid units are added to the end of a growing protein molecule. 4. Genetic material must be capable of undergoing occasional mutations in which the information it carries is altered. If these mutations are to be heritable, the mutant molecules must also be capable of being replicated as faithfully as the parental molecule. This feature is necessary to account for the evolution of diverse organisms through the slow accumulation of favorable mutations. Watson and Crick suggested that heritable mutations might be possible in DNA by rare mispairing of the bases, with the result that an incorrect nucleotide becomes incorporated into a replicating DNA strand.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

SUMMING UP ■ DNA is a helical molecule, consisting of two antiparallel strands. ■ The two strands of DNA are complementary: A pairs with T, and G pairs with C. ■ The helix is stabilized by the hydrogen bonds that form between complementary bases, as well as by the stacking interactions that occur between adjacent base pairs. ■ The structure of DNA is consistent with the four necessary properties of genetic material.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

ROOTS OF DISCOVERY The Double Helix James D. Watson and Francis H. C. Crick (1953) Cavendish Laboratory Cambridge, England

A Structure for Deoxyribose Nucleic Acid This is one of the watershed papers of twentieth-century biology. Watson and Crick benefited tremendously from knowing that their structure was consistent with the unpublished structural studies of Maurice Wilkins and Rosalind Franklin. The same issue of Nature that included the Watson and Crick paper also included, back to back, a paper from the Wilkins group and one from the Franklin group detailing their data and the consistency of their data with the proposed structure. It has been said that Franklin was poised a mere two half-steps from making the discovery herself, alone. In any event, Watson and Crick and Wilkins were awarded the 1962 Nobel Prize for their discovery of DNA structure. Rosalind Franklin, tragically, died of cancer in 1958 at the age of 38.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

“If only specific pairs of bases can be formed, it follows that if the sequence of bases on one chain is given, then the sequence

”

on the other chain is automatically determined.

Watson and Crick's 1953 paper was concise, occupying only a single printed page in the journal Nature. It describes the basic molecular model—the helical structure, the anti-parallel strands, and the physical positioning of the bases and the phosphate groups. It then describes base pairing and its significance.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Only specific pairs of bases can bond together. These pairs are adenine (purine) with thymine (pyrimidine), and guanine (purine) with cytosine (pyrimidine)…. The sequence of bases on a single chain does not appear to be restricted in any way. However, … it follows that if the sequence of bases on one chain is given, then the sequence on the other chain is automatically determined…. It has not escaped our notice that the specific pairing we have postulated immediately suggests a plausible copying mechanism for the genetic material. This last sentence is one of the most widely quoted lines in all of biology. In a subsequent, more detailed paper, Watson and Crick went a step further and considered how information could be encoded in DNA, given their proposed molecular structure. The phosphate–sugar backbone of our model is completely regular, but any sequence of the pairs of bases can fit into the structure. It follows that in a long molecule many different permutations are possible, and it therefore seems likely that the precise sequence of the bases is the code which carries the genetical information.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

While Watson and Crick's first paper is considered iconic, it is in the second one that the biological and genetic significance of the double helix is most fully explored. Source: J.D. Watson and F.H.C. Crick, Molecular Structure of Nucleic Acids: A Structure for Deoxyribose Nucleic Acid. Nature 171 (1953): 737–738.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.4 The Separation and Identification of Genomic DNA Fragments The following sections show how an understanding of DNA structure and replication has been put to practical use in the development of procedures for the separation and identification of particular DNA fragments. These methods are used primarily either to identify DNA markers or to aid in the isolation of particular DNA fragments that are of genetic interest. For example, consider a pedigree of familial breast cancer in which a particular DNA fragment serves as a marker for a region of chromosome that also includes the gene, an allele of which is responsible for the increased risk. In this case, the ability to identify the marker genotype for each woman in the pedigree is critically important for predicting her risk of breast cancer. To take another example, suppose one is testing the hypothesis that an allelic variant associated with a hereditary disease is present in a particular DNA fragment. In this situation it is important to be able to pinpoint this fragment using genetic markers so as to isolate the fragment from affected individuals, verify whether the hypothesis is true, and identify the nature of the mutation.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Most procedures for the separation and identification of DNA fragments can be grouped into two general categories: ■ Those that identify a specific DNA fragment present in genomic DNA by making use of the fact that complementary singlestranded DNA sequences can, under the proper conditions, form a duplex molecule. These procedures rely on nucleic acid hybridization. ■ Those that use prior knowledge of the sequence at the ends of a DNA fragment to specifically and repeatedly replicate this one fragment from genomic DNA. These procedures rely on

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

selective DNA replication (amplification) by means of the polymerase chain reaction. The major difference between these approaches is that the first (relying on nucleic acid hybridization) identifies fragments that are present in the genomic DNA itself, whereas the second (relying on DNA amplification) identifies experimentally manufactured replicas of fragments whose original templates (but not the replicas) were present in the genomic DNA. This difference has practical implications: ■ Hybridization methods require a greater amount of genomic DNA for the experimental procedures, but relatively large fragments can be identified, and no prior knowledge of the DNA sequence is necessary.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Amplification methods require extremely small amounts of genomic DNA for the experimental procedures, but the amplification is usually restricted to relatively small fragments, and some prior knowledge of DNA sequence is necessary. Historically, hybridization methods were the most important means for detecting specific DNA fragments. However, with the development of the polymerase chain reaction and automated DNA sequencing technologies in the late 1980s, amplification-based methods came to predominate these investigation. In the following section, we will briefly describe hybridization methods, after which we will focus in more depth on amplification-based methods.

Restriction Enzymes and Site-Specific DNA Cleavage In methods that use nucleic acid hybridization to identify particular fragments present in genomic DNA, the first step is usually cutting the genomic DNA into specific fragments of experimentally

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

manageable size. When genomic DNA is isolated from cells, it is typically fragmented into pieces with an average length of about 50,000 bases, or 50 kilobases (kb). It is possible to fragment them further, but the breakage is by and large random. Thus, only a small fraction of all of the fragments will contain a particular DNA sequence, and those fragments will vary in size (FIGURE 2.11). This creates a “needle-in-a-haystack” problem: How can that sequence of interest be separated from all of the rest?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.11 Illustration of random fragments of DNA, some of which carry a sequence of interest (shown in red).

One of the most important discoveries in the history of molecular genetics (for which Werner Arbers, Daniel Nathans, and Hamilton Smith shared the Nobel Prize in 1978) was that of restriction endonucleases or restriction enzymes. These enzymes, which are found in bacteria, can cleave DNA molecules at positions at which specific short sequences of DNA (typically 4–6 base pairs in length) occur. Restriction enzymes function in nature to protect bacteria by selectively degrading the genomes of bacteriophages that attack them. For example, the restriction enzyme BamH1 (technically known as a type II restriction endonuclease) recognizes the sequence

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

and cleaves each strand between the two G-bearing nucleotides, as shown in FIGURE 2.12.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.12 The mechanism of DNA cleavage by the restriction enzyme BamHI. Wherever the duplex contains a BamHI restriction site, the enzyme makes a single cut in the backbone of each DNA strand. Each cut creates a new 3′ end and a new 5′ end, separating the duplex into two fragments. In the case of BamHI, the cuts are staggered cuts, so the resulting ends terminate in single-stranded regions, each four nucleotides in length.

FIGURE 2.13 shows nine of the several hundred restriction enzymes that are known. Most restriction enzymes are named after the species in which they were found. BamHI, for example, was isolated from the bacterium Bacillus amyloliquefaciens strain H, and it is the first (I) restriction enzyme isolated from this organism. Because the first three letters in the name of each restriction enzyme stand for the bacterial species of origin, these

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

letters are printed in italics; the rest of the symbols in the name are not italicized.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.13 Recognition sites for various restriction enzymes. The vertical dashed line indicates the axis of symmetry in each sequence. Red arrows indicate the sites of cutting. Enzymes in the first two columns produce sticky ends; those in the third column produce blunt ends. The enzyme TaqI yields cohesive ends consisting of two nucleotides, whereas the cohesive ends produced by the other enzymes contain four nucleotides. R and Y refer to any complementary purines and pyrimidines, respectively.

Most restriction enzymes recognize only one short base sequence, usually four or six nucleotide pairs. The enzyme binds with the DNA at these sites and makes a break in each strand of the DNA molecule, producing free 3′-OH and 5′-P groups at each position. The nucleotide sequence recognized for cleavage by a restriction enzyme is called the restriction site of the enzyme. Some restriction enzymes cleave their restriction site asymmetrically (at different sites in the two DNA strands), but other restriction enzymes cleave the site symmetrically (at the same site in both

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

strands). The former leave sticky ends (also referred to as cohesive ends) because each end of the cleaved site has a small, single-stranded overhang that is complementary in base sequence to the other end (Figure 2.12). In contrast, enzymes that have symmetrical cleavage sites yield DNA fragments that have blunt ends. In virtually all cases, the restriction site of a restriction enzyme reads the same on both strands, provided that the opposite polarity of the strands is taken into account; for example, each strand in the restriction site of BamHI reads 5′-GGATCC-3′

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(Figure 2.10). A DNA sequence with this type of symmetry is called a palindrome. (In ordinary English, a palindrome is a word or phrase that reads the same forward and backward, such as “madam.”)

Restriction enzymes have the following important characteristics: ■ Most restriction enzymes recognize a single restriction site. ■ The restriction site is recognized without regard to the source of the DNA. ■ Because most restriction enzymes recognize a unique restriction-site sequence, the number of cuts in the DNA from a

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

particular organism is determined by the number of restriction sites present. The DNA fragment produced by a pair of adjacent cuts in a DNA molecule is called a restriction fragment. A large DNA molecule typically will be cut into many restriction fragments of different sizes. For example, an E. coli DNA molecule, which contains 4.6 × 106 base pairs, is cut into several hundred to several thousand

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

fragments, and mammalian genomic DNA is cut into more than a million fragments. Most importantly, these fragments are not generated randomly; instead, their ends are determined by the presence of restriction sites in the DNA being digested. Therefore, in digested DNA, a sequence of interest should always occur in fragments with identical ends. This has three important implications: 1. Restriction sites can be used as genetic markers if sequence variation within particular sites exists in a population of individuals. 2. If a particular restriction fragment can be isolated, then the sequences contained within it can be further characterized. 3. Two fragments of DNA, generated by digestion with an enzyme like BamH1, will have complementary ends that can (at least in theory) base-pair with each other. In fact, this base pairing can be readily accomplished, and it forms the basis for cloning of DNA shown in Figure 2.4 and described in the Manipulating Genes and Genomes chapter.

Gel Electrophoresis So if we now have genomic DNA that has been digested, how do we separate the thousands of different fragments? They can be separated by size using the fact that DNA is negatively charged and moves in response to an electric field. If the terminals of an

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

electrical power source are connected to the opposite ends of a horizontal tube containing a DNA solution, then the DNA molecules will move toward the positive end of the tube at a rate that depends on the electric field strength and on the shape and size of the molecules. The movement of charged molecules in an electric field is called electrophoresis.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The type of electrophoresis most commonly used in genetics is gel electrophoresis. An experimental arrangement for gel electrophoresis of DNA is shown in FIGURE 2.14A. A thin slab of a gel, usually agarose or acrylamide, is prepared containing small slots (called wells) into which samples are placed. An electric field is applied, and the negatively charged DNA molecules penetrate and move through the gel toward the anode (the positively charged electrode). A gel is a complex molecular network that contains narrow, tortuous passages, so smaller DNA molecules pass through more easily; hence the rate of movement increases as the size of the DNA fragment decreases. FIGURE 2.14B shows the result of electrophoresis of a set of double-stranded DNA molecules in an agarose gel. Each discrete region containing DNA is called a band. The bands can be visualized under ultraviolet light after soaking the gel in the dye ethidium bromide, the molecules of which intercalate into duplex DNA and render it fluorescent. In Figure 2.14B, each band in the gel results from the fact that all DNA fragments of a given size have migrated to the same position in the gel. To produce a visible band, a minimum of about 5 × 10−9 grams of DNA is required, which for a fragment of size 3 kb works out to approximately 109 molecules. The point is that a very large number of copies of any particular DNA fragment must be present to yield a visible band in an electrophoresis gel.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.



Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.14 Gel electrophoresis of DNA. (A) Liquid gel is allowed to harden with an appropriately shaped mold in place to form slots for the samples. After electrophoresis, the DNA fragments, which are located at various positions in the gel, are made visible by immersing the gel in a solution containing a reagent that binds to or reacts with DNA. (B) After staining with a fluorescent dye (ethidium bromide), the separated fragments in a sample appear as bands when the gel is exposed to ultraviolet light.

Because of the sequence specificity of cleavage, a particular restriction enzyme produces a unique set of fragments for a particular DNA molecule. Another enzyme will produce a different

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

set of fragments from the same DNA molecule. In FIGURE 2.15, this principle is illustrated for the digestion of a circular molecule of double-stranded DNA with a length of 10 kb. When digested with the restriction enzyme EcoRI (Figure 2.15A), the circular molecule yields bands of 4 kb and 6 kb. This pattern would result from EcoRI restriction sites located in the circle at the relative positions shown beneath the gel. The circle is oriented arbitrarily, with one of the EcoRI (E) sites at the top. Similarly, digestion of the circle with the enzyme BamHI (Figure 2.15B) results in bands of 3 kb and 7

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

kb, which implies that the circle contains BamHI sites at the positions indicated in the diagram beneath the gel. Again the circle is oriented arbitrarily, this time with one of the BamHI (B) sites located at the top. A diagram showing sites of cleavage of one or more restriction sites along a DNA molecule is called a restriction map.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.15 Gel diagrams showing the sizes of restriction fragments produced by digestion of a 10-kb circular molecule of double-stranded DNA with (A) EcoRI, (B) BamHI, and (C) both enzymes together. Beneath each diagram is a restriction map of the circular DNA showing the locations of the restriction sites. The restriction map in (C) takes into account those in (A) and (B) as well as the fragment sizes produced by digestion with both enzymes.

When both EcoRI and BamHI are used together, the resulting DNA fragments reveal where the EcoRI sites and the BamHI sites are located relative to each other. In this case, digestion with both enzymes yields bands of 1 kb, 2 kb, 3 kb, and 4 kb (Figure 2.15C). The restriction map shown beneath the gel indicates where the two types of restriction sites must be located to yield these band sizes. This restriction map can be obtained by superimposing that in part B over that in part A, and rotating the result until the distances between adjacent pairs of restriction sites equal 1, 2, 3, and 4 kb (not necessarily in that order). In this case, one need only

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

rotate the restriction map in part B a distance of 2 kb to the right. Note that, in the restriction-enzyme digest in Figure 2.13C, the 4kb fragment is not the same 4-kb fragment as observed in part A, and the 3-kb fragment is not the same 3-kb fragment as observed in part B. This discordance arises because each restriction enzyme cleaves the fragments produced by the other. The orientation of the restriction map in Figure 2.13C is arbitrary. It can be flipped over or rotated by any amount in any direction, and it will still be the same restriction map. Construction of such a map facilitates two outcomes. First, the individual fragments can be isolated from the gel, inserted into selfreplicating molecules such as bacteriophage, plasmids, or even small artificial chromosomes (Figure 2.4), and introduced into bacterial cells (DNA cloning, a topic we will cover in the Manipulating Genes and Genomes chapter). Second, and most important with respect to DNA markers, we have taken the first step toward identifying particular positions in DNA—those at which the enzymes cleave. The question thus becomes how we can apply restriction analysis to complex genomes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Nucleic Acid Hybridization Most genomes are sufficiently large and complex that digestion with a restriction enzyme produces many bands that are the same or similar in size. Identifying a particular DNA fragment within a background of many other fragments of similar size presents a needle-in-a-haystack problem. Suppose, for example, that we are interested in a particular 3.0-kb BamHI fragment from the human genome that serves as a marker indicating the presence of a genetic risk factor for breast cancer among women in a particular pedigree. On the basis of size alone, this fragment of 3.0 kb is indistinguishable from fragments ranging in size from about 2.9 to

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

3.1 kb. How many fragments in this size range are expected? When human genomic DNA is cleaved with BamHI, the average length of a restriction fragment is 46 = 4096 base pairs, and the expected total number of BamHI fragments is about 730,000; in the size range 2.9–3.1 kb, the expected number of fragments is about 17,000. Thus, even though we know that the fragment of interest is 3 kb in length, it is only one of 17,000 fragments that are so similar in size that the sought-after fragment cannot be distinguished from the others by length alone. In fact, as seen in FIGURE 2.16,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

digested genomic DNA, when visualized, appears as a smear rather than as a set of discrete bands.

FIGURE 2.16 An example of genomic DNA from various plants digested with the enzymes EcoR1 and HindIII. Reproduced from Kang, T. J., & Yang, M. S. Rapid and reliable extraction of genomic DNA from various wild-type and transgenic plants. BMC Biotechnol. 2004;4:20. doi:10.1186/1472-6750-4-20. Creative Commons license available: https://creativecommons.org/licenses/by/2.0/

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

This identification task is actually more difficult than finding a needle in a haystack because haystacks are usually dry. A more accurate analogy would be looking for a needle in a haystack that had been pitched into a swimming pool full of water. This analogy is more relevant because gels, even though they contain a supporting matrix to make them semisolid, are primarily composed of water, and each DNA molecule within a gel is surrounded entirely by water. Clearly, we need some method by which specific fragments can be either visualized or purified.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

At this point, we need to return to the structure of DNA (Figure 2.8), examine how the two strands in a double helix can be separated to form single strands, and see how, under the proper conditions, two single strands that are complementary or nearly complementary in sequence can come back together to form a different double helix. The separation of the strands is called denaturation, and the coming together of complementary strands is called nucleic acid hybridization or renaturation. (The term hybridization is appropriate because the two strands that anneal to form a DNA duplex may not be exactly the same strands that were paired prior to denaturation.) The practical applications of nucleic acid hybridization are many: ■ A small part of a DNA fragment can be hybridized with a much larger DNA fragment. This principle is used in identifying specific DNA fragments in a complex mixture, such as the 3-kb BamHI marker for breast cancer that we have been considering. Applications of this type include tracking of genetic markers in pedigrees and isolation of fragments containing a particular mutant gene. ■ A DNA fragment from one gene can be hybridized with similar fragments from other genes in the same genome; this principle is used to identify different members of families of genes that

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

are similar, but not identical, in sequence and that have related functions. ■ A DNA fragment from one species can be hybridized with similar sequences from other species. This allows the isolation of genes that have the same or related functions in multiple species. This method is used to study aspects of molecular evolution, such as how differences in sequence are correlated with differences in function, and the patterns and rates of change in gene sequences as they evolve.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

As we saw in Section 2.3, The Molecular Structure of DNA, the double-stranded helical structure of DNA is maintained by base stacking and by hydrogen bonding between the complementary base pairs. When solutions containing DNA fragments are raised to temperatures in the range 85°C–100°C, or to the high pH of strong alkaline solutions, the paired strands begin to separate. Unwinding of the helix happens in less than a few minutes (the time depends on the length of the molecule). A common way to detect denaturation is by measuring the capacity of DNA in solution to absorb ultraviolet light of wavelength 260 nm, because the absorption at 260 nm (A260) of a solution of single-stranded molecules is 37 percent higher than the absorption of the doublestranded molecules at the same concentration. As shown in FIGURE 2.17, the progress of denaturation can be followed by slowly heating a solution of double-stranded DNA and recording the value of A260 at various temperatures. The temperature required for denaturation increases with G + C content, not only because G–C base pairs have three hydrogen bonds and A–T base pairs have two such bonds, but also because consecutive G–C base pairs have stronger base stacking.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE 2.17 The mechanism of denaturation of DNA by heat. The temperature at which 50 percent of the base pairs are denatured is the melting temperature, symbolized as Tm.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

For denatured DNA strands to be able to undergo renaturation, two requirements must be met: 1. The salt concentration must be high (greater than 0.25 M) to neutralize the negative charges of the phosphate groups, which would otherwise cause the complementary strands to repel each other. 2. The temperature must be high enough to disrupt hydrogen bonds that form at random between short sequences of bases within the same strand, but not so high that stable base pairs between the complementary strands are disrupted. The initial phase of renaturation is a slow process because the rate is limited by the chance that a region of two complementary strands will come together at random to form a short sequence of correct base pairs. This initial pairing step is followed by a rapid pairing of the remaining complementary bases and rewinding of the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

helix. Rewinding is accomplished in a matter of seconds, and its rate is independent of DNA concentration because the complementary strands have already found each other. Importantly, if two DNA molecules from different sources contain identical sequences and are allowed to reanneal together, then some of the renatured DNA will consist of heteroduplex DNA— that is, molecules containing base-paired strands that originated from different sources (FIGURE 2.18). If one DNA source is genomic DNA, and the other is a DNA probe, consisting of part of our sequence of interest that is suitably labeled (for example, with radioactive 32P), then all of the heteroduplexes created will be

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

radioactively labeled. Geneticists say that probe DNA hybridizes with DNA fragments containing sequences that are similar, rather than complementary to the probe.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.18 DNA detection by nucleic acid hybridization. Duplex genomic DNA is denatured with heat, after which it is combined with a labeled probe sequence homologous to a specific sequence in the genomic DNA. The mixture is cooled, allowing formation of heteroduplex molecules containing one strand of the genomic DNA base-paired with the labeled probe sequence.

This process can be used to address the problem of visualizing particular fragments in a conceptually simple way. If denatured DNA is immobilized on a membrane (typically made of nitrocellulose or nylon membrane), then that membrane can be incubated in a solution containing a radioactively labeled probe, consisting of copies of the sequence of interest. This process, which is shown in FIGURE 2.19, is referred to as a Southern blot (named after E. M. Southern, who developed the technique). In this process, digested genomic DNA is transferred from an agarose gel to a

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

membrane. DNA on the membrane is then denatured, after which it is incubated with a denatured probe under conditions that allow renaturation and heteroduplex formation. The pattern of hybridization can then be visualized by exposure to x-ray film.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.19 The Southern blot: an experimental procedure for identifying the position of a specific DNA fragment in a gel.

Returning to our original problem, that of testing DNA for a marker of a breast cancer risk allele in a pedigree, we could use this technique to analyze DNA from all of the individuals from whom we could obtain genomic DNA. In fact, this process is extremely sensitive: Under typical conditions, a band can be observed on the film that contains only 5 × 10−12 grams of DNA—a thousand times less DNA than the amount required to produce a visible band in the gel itself.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

SUMMING UP ■ Restriction enzymes cleave DNA at specific recognition sites, generating specific fragments. ■ These fragments can be analyzed using gel electrophoresis. ■ Restriction sites can be mapped on DNA molecules, and restriction fragments can be isolated for further study. ■ DNA can be denatured and reannealed.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Restriction analysis of genomic DNA can be performed by hybridizing electrophoretically separated DNA with labeled specific probes.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.5 Amplification of Specific DNA for Detection and Purification Despite its sensitivity, Southern blotting has its limitations. It requires a relatively large amount of genomic DNA, and in some applications (for example, analysis of trace amounts of DNA at a crime scene or from ancient remains), that kind of sample is not

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

available. In addition, while it allows a particular DNA fragment to be identified when present in a complex mixture of fragments, it does not enable the fragment to be separated from the others and purified. Obtaining the fragment in purified form requires cloning, which is straightforward but time consuming. (Cloning methods are discussed in the Manipulating Genes and Genomes chapter.) However, if the fragment of interest is not too long, and if the nucleotide sequence at each end is known, then it becomes possible to obtain large quantities of the fragment merely by selective replication. This process is called amplification. Once this is accomplished, the amplified fragment can be analyzed directly. In fact, these procedures have largely supplanted hybridization methods in genetic screening protocols. How would one know the nucleotide sequences at the ends of interest? Let us return to our example of the 3.0-kb BamHI fragment that serves to mark a risk allele for breast cancer in certain pedigrees. Suppose that this fragment is cloned and sequenced from one affected individual, and it is found that, relative to the sequence in unaffected individuals, the BamHI fragment is missing a region of 500 base pairs. At this point, the sequences at the ends of the fragment are known, and we can also infer that amplification of genomic DNA from individuals with the risk factor will yield a band of 3.0 kb, whereas amplification from genomic DNA of noncarriers will yield a band of 3.5 kb. This difference allows every person in the pedigree to be diagnosed as a carrier or

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

noncarrier merely by means of DNA amplification. To understand how amplification works, it is first necessary to examine a few key features of DNA replication.

Constraints on DNA Replication: Primers and 5′-to-3′ Strand Elongation

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

As was the case with restriction analysis, amplification utilizes a biological process—in this case, DNA replication. In doing so, it uses the enzyme that forms the sugar–phosphate bond (the phosphodiester bond) between adjacent deoxynucleotides in a DNA chain, called a DNA polymerase. A variety of DNA polymerases have been purified, and for amplification of a DNA fragment, the DNA synthesis is carried out in vitro by combining purified cellular components in a test tube under precisely defined conditions. (In vitro, literally “in glass,” implies the absence of living cells.) For DNA polymerase to catalyze synthesis of a new DNA strand, preexisting single-stranded DNA must be present. Each singlestranded DNA molecule present in the reaction mix can serve as a template upon which a new partner strand is created by the DNA polymerase. For DNA replication to take place, the 5′-triphosphates of the four deoxynucleoside triphosphates must also be present. This requirement is rather obvious, because these are the precursors from which new DNA strands are created. The triphosphates needed are the compounds denoted in Table 2.1 as dATP, dGTP, dTTP, and dCTP, which contain the bases adenine, guanine, thymine, and cytosine, respectively. Details of the structures of dCTP and dGTP are shown in FIGURE 2.20, in which the phosphate groups cleaved off during DNA synthesis are indicated. DNA synthesis requires all four nucleoside 5′triphosphates and does not take place if any of them are omitted.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.20 Two deoxynucleoside triphosphates used in DNA synthesis. The outer two phosphate groups are removed during synthesis.

One feature of all DNA polymerases is that a DNA polymerase can only elongate a DNA strand. It is not possible for DNA polymerase to initiate synthesis of a new strand, even when a template molecule is present. An important implication of this principle is that DNA synthesis requires a preexisting segment of nucleic acid that is hydrogen-bonded to the template strand; this segment is called a primer. Because the primer molecule can be very short, it is an oligonucleotide, which literally means “few nucleotides.” As we shall see in the DNA Replication and Sequencing chapter, in living cells the primer is a short segment of RNA; in contrast, in DNA amplification in vitro, the primer employed is usually DNA. It is the 3′ end of the primer that is essential, because DNA synthesis proceeds only by addition of successive nucleotides to the 3′ end of the growing strand. In other words, chain elongation always takes place in the 5′-to-3′ direction (5′ → 3′). The reason

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

for the 5′ → 3′ direction of chain elongation is illustrated in FIGURE 2.21: The reaction catalyzed by DNA polymerase is the formation of a phosphodiester bond between the free 3′-OH group of the chain being extended and the innermost phosphorus atom of the nucleoside triphosphate being incorporated at the 3′ end. Recognition of the appropriate incoming nucleoside triphosphate in replication depends on base pairing with the opposite nucleotide in the template strand. DNA polymerase will usually catalyze the polymerization reaction that incorporates the new nucleotide at the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

primer terminus only when the correct base pair is present. The same DNA polymerase is used to add each of the four deoxynucleoside phosphates to the 3′-OH terminus of the growing strand.

FIGURE 2.21 Addition of nucleotides to the 3′-OH terminus of a growing strand. The recognition step is shown as the formation of hydrogen bonds between the G and the C. The chemical reaction is that the 3′-OH group of the 3′ end of the growing chain attacks the innermost phosphate group of the incoming trinucleotide.

The Polymerase Chain Reaction

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

The requirement for an oligonucleotide primer, and the constraint that chain elongation must always occur in the 5′ → 3′ direction, make it possible to obtain large quantities of a particular DNA sequence by selective amplification in vitro. The method for selective amplification is called the polymerase chain reaction

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(PCR). For its invention, Californian Kary B. Mullis was awarded a Nobel Prize in 1993. PCR amplification uses DNA polymerase and a pair of short, synthetic oligonucleotide primers, usually 18–22 nucleotides in length, that are complementary in sequence to the ends of the DNA sequence to be amplified. FIGURE 2.22 gives an example in which the primer oligonucleotides (green) are 9-mers. These are too short for most practical purposes, but they will serve for illustration. The original duplex molecule (part A) is shown in blue. This duplex is mixed with a vast excess of primer molecules, DNA polymerase, and all four nucleoside triphosphates. When the temperature is raised, the strands of the duplex denature and become separated. When the temperature is lowered again to allow renaturation, the primers, because they are in great excess, hybridize (or anneal) to the separated template strands (part B). Note that the primer sequences are different from each other but complementary to sequences present in opposite strands of the original DNA duplex and flanking the region to be amplified. The primers are oriented with their 3′ ends pointing in the direction of the region to be amplified, because each DNA strand elongates only at the 3′ end. After the primers have annealed, each is elongated by DNA polymerase using the original strand as a template, and the newly synthesized DNA strands (red) grow toward each other as synthesis proceeds (part C). Note that a region of duplex DNA present in the original reaction mix can be PCR-amplified only if the region is flanked by the primer oligonucleotides.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.22 Role of primer sequences in PCR amplification. (A) Target DNA duplex (blue), showing sequences chosen as the primer-binding sites flanking the region to be amplified. (B) Primer (green) bound to denatured strands of target DNA. (C) First round of amplification. Newly synthesized DNA is shown in pink. Note that each primer is extended beyond the other primer site. (D) Second round of amplification (only one strand shown); in this round, the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

newly synthesized strand terminates at the opposite primer site. (E) Third round of amplification (only one strand shown); in this round, both strands are truncated at the primer sites. Primer sequences are normally at least twice as long as shown here.

To start a second cycle of PCR amplification, the temperature is raised again to denature the duplex DNA. Upon lowering of the temperature, the original parental strands anneal with the primers

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

and are replicated as shown in Figure 2.22B and C. The daughter strands produced in the first round of amplification also anneal with primers and are replicated, as shown in part D. In this case, although the daughter duplex molecules are identical in sequence to the original parental molecule, they consist entirely of primer oligonucleotides and nonparental DNA that was synthesized in either the first or the second cycle of PCR. As successive cycles of denaturation, primer annealing, and elongation occur, the original parental strands are diluted out by the proliferation of new daughter strands until eventually almost every molecule produced in the PCR has the structure shown in part E. The power of PCR amplification derives from the fact that the number of copies of the template strand increases in exponential progression: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, and so forth, doubling with each cycle of replication. Starting with a mixture containing as little as one molecule of the fragment of interest, repeated rounds of DNA replication increase the number of amplified molecules exponentially. For example, starting with a single molecule, 25 rounds of DNA replication will result in 225 = 3.4 × 107 molecules. This number of molecules of the amplified fragment is so much greater than that of the other unamplified molecules in the original mixture that the amplified DNA can often be used without further purification. For example, a single 3-kb

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

fragment in E. coli accounts for only 0.06 percent of the DNA in this organism. However, if this single fragment were replicated through 25 rounds of replication, then 99.995 percent of the resulting mixture would consist of the amplified sequence. A 3-kb fragment of human DNA constitutes only 0.0001 percent of the total genome size. Amplification of a 3-kb fragment of human DNA to 99.995 percent purity would require approximately 34 cycles of PCR. FIGURE 2.23 provides an overview of the polymerase chain

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

reaction. The DNA sequence to be amplified is again shown in blue and the oligonucleotide primers in green. The oligonucleotides anneal to the ends of the sequence to be amplified and become the substrates for chain elongation by DNA polymerase. In the first cycle of PCR amplification, the DNA is denatured to separate the strands. The denaturation temperature is usually around 95°C. The temperature is then decreased to allow annealing in the presence of a vast excess of the primer oligonucleotides. The annealing temperature is typically in the range of 50°C–60°C, depending largely on the G + C content of the oligonucleotide primers. To complete the cycle, the temperature is raised slightly, to about 70°C, for the elongation of each primer. The steps of denaturation, renaturation, and replication are repeated from 20 to 30 times, and in each cycle the number of molecules of the amplified sequence is doubled.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.23 Polymerase chain reaction (PCR) for amplification of particular DNA sequences. Oligonucleotide primers (green) that are complementary to the ends of the target sequence (blue) are used in repeated rounds of denaturation, annealing, and DNA replication. Newly replicated DNA is shown in pink. The number of copies of the target sequence doubles in each round of replication, eventually overwhelming any other sequences that may be present.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Implementation of PCR with conventional DNA polymerases is not practical, because at the high temperature necessary for denaturation, the polymerase is itself irreversibly unfolded and becomes inactive. However, DNA polymerase isolated from certain types of organisms is heat stable because those organisms normally live in hot springs at temperatures well above 90°C, such as are found in Yellowstone National Park. Such organisms are said to be thermophiles. The most widely used heat-stable DNA polymerase is called Taq polymerase, because it was originally

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

isolated from the thermophilic bacterium Thermus aquaticus. PCR amplification is very useful for generating large quantities of a specific DNA sequence. The principal limitation of the technique is that the DNA sequences at the ends of the region to be amplified must be known so that primer oligonucleotides can be synthesized. In addition, sequences longer than about 5000 base pairs cannot be replicated efficiently by conventional PCR procedures, although some modifications of PCR allow longer fragments to be amplified. Conversely, many applications require amplification of relatively small fragments. The major advantage of PCR amplification is that it requires only trace amounts of template DNA. Theoretically, only one template molecule is required, but in practice the amplification of a single molecule may fail because the molecule may, by chance, be broken or damaged. In practice, amplification is usually reliable with as few as 10–100 template molecules, which makes PCR amplification 10,000–100,000 times more sensitive than detection via nucleic acid hybridization. With PCR, genotyping individuals is greatly simplified. Rather than having to go through all the steps involved in Southern blotting, a small amount of genomic DNA can be used in combination with appropriate primers to amplify the sequence of interest. Returning to our example of a breast cancer risk marker, FIGURE 2.24

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

illustrates what we would observe if we genotyped two individuals, one homozygous and one heterozygous for the allele of interest. Furthermore, the exquisite sensitivity of PCR amplification has led to its use in DNA typing for criminal cases in which a minuscule amount of biological material has been left behind by the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

perpetrator (skin cells on a cigarette butt or hair-root cells on a single hair can yield enough template DNA for amplification).

FIGURE 2.24 Determining a DNA marker genotype based on PCR. Shown are the genotypes of two individuals. Individual 1 is homozygous; individual 2 is heterozygous for two alleles that differ in length, due to a deletion in the second allele (shown in orange). The PCR products can be directly analyzed without resorting to blotting, resulting in one band from individual 1 and two bands from individual 2.

In research, PCR is widely used in the study of independent mutations in a gene whose sequence is known so as to identify the molecular basis of each mutation, to study DNA sequence

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

variations among alternative forms of a gene that may be present in natural populations, or to examine differences among genes with the same function in different species. The PCR procedure has also come into widespread use in clinical laboratories for diagnosis. To take just one very important example, the presence of the human immunodeficiency virus (HIV), which causes acquired immunodeficiency syndrome (AIDS), can be detected in trace quantities in blood banks via PCR by using primers complementary to sequences in the viral genetic material. These and other applications of PCR are facilitated by the fact that the procedure lends itself to automation by the use of mechanical robots to set up and run the reactions.

SUMMING UP ■ Specific DNA fragments contained in a complex mixture like genomic DNA can be selectively amplified with the polymerase chain reaction (PCR).

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ PCR requires the presence of template DNA, nucleoside triphosphates, DNA polymerase, and two oligonucleotide primers that flank the region to be amplified. ■ PCR requires repeated rounds of DNA synthesis, denaturation, and primer reannealing. This is accomplished using a thermocycler and a heat-stable DNA polymerase. ■ Use of PCR greatly simplifies the genotyping of DNA markers, and it can be used for many other applications, including pathogen detection and forensic DNA analysis.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.6 Types of DNA Markers Present in Genomic DNA Genetic variation, in the form of polymorphisms, exists in most natural populations of organisms. The methods of DNA manipulation examined in Sections 2.4 and 2.5 can be used in a variety of combinations to detect differences among individuals. Anyone who reads the literature in modern genetics will encounter a bewildering variety of acronyms referring to different ways in which genetic polymorphisms are detected. The different approaches are in use because no single method is ideal for all applications, each method has its own advantages and limitations, and new methods are continually being developed. In this section, we examine some of the principal methods for detecting DNA polymorphisms among individuals.

Restriction Fragment Length

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Polymorphisms The simplest kind of genetic polymorphism is a single-nucleotide polymorphism (SNP) (pronounced “snip”). This is simply a position in the genome where, within a population, two or more different bases are found in different genomes. We now know that SNPs are ubiquitous: The 1000 Genomes Project has cataloged more than 80 million of them in the human genome. However, prior to the development of modern high-throughput genotyping, detection of SNPs and other forms of sequence variation was difficult. Restriction enzymes proved to be valuable tools in elucidating their presence. We have already seen that restriction enzymes can be used to map the occurrence of particular sequences (the recognition sites) in DNA molecules. If a recognition site contains a DNA polymorphism,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

then digestion can be used as a means of detecting it. An example is shown in FIGURE 2.25, in which a T–A nucleotide pair appears in some molecules and a C–G pair in others. In this example, the polymorphic nucleotide site is included in a cleavage site for the restriction enzyme EcoRI (5′-GAATTC-3′). The two nearest flanking EcoRI sites are also shown. In this kind of situation, DNA molecules with T–A nucleotide pairs will be cleaved at both flanking sites as well as at the middle site, yielding two EcoRI restriction fragments. In contrast, DNA molecules with C–G nucleotide pairs will be

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

cleaved at both flanking sites but not at the middle site (because the presence of C–G destroys the EcoRI restriction site), so they will yield only one larger restriction fragment. A SNP that eliminates a restriction site is known as a restriction fragment length polymorphism (RFLP) (pronounced either as “riflip” or by spelling it out).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.



FIGURE 2.25 A difference in the DNA sequence of two molecules can be detected if the difference eliminates a restriction site. (A) This molecule contains three restriction sites for EcoRI, including one at each end. It is cleaved into two fragments by the enzyme. (B) This molecule has an altered EcoRI site in the middle, in which 5′-GAATTC-3′ becomes 5′-GAACTC-3′. The altered site cannot be cleaved by EcoRI, so treatment of this molecule with EcoRI results in one larger fragment.

In the early days of analysis of genomic DNA, prior to the development of automated DNA sequencing (see the DNA Replication and Sequencing chapter), RFLP analysis was an extremely important tool for mapping chromosomes and detecting

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

DNA polymorphisms, since such polymorphisms could be detected by Southern blotting. With the advent of PCR, however, the process could be greatly simplified. As shown in FIGURE 2.26, primers that flank the RFLP site can be used to amplify the DNA containing it, and then digestion and simple gel electrophoresis can

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

be used to analyze the product. Given that a fragment that has been amplified can also be sequenced, comparison of products from different alleles can then be extended to include all positions in the fragment.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE 2.26 In a restriction fragment length polymorphism (RFLP), alleles may differ in the presence or absence of a cleavage site in the DNA. This example shows a PCR fragment containing two single-base-pair alleles. Primer sequences are shown in blue. The a allele lacks a restriction site that is present in the DNA of the A allele. The difference in fragment length can be detected by digestion of PCR products containing the cleavage site. RFLP alleles are codominant, which means that DNA from the heterozygous Aa genotype yields each of the single bands observed in DNA from homozygous AA and aa genotypes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Single-Nucleotide Polymorphisms RFLP analysis can detect only those allelic variants that affect the pattern of digestion of the enzyme(s) used, which are only a small fraction of the vast number of allelic differences that exist in genomic DNA. If, instead of digesting the fragments, their sequences are determined completely, then SNPs can be detected. A SNP is a particular nucleotide site in DNA that differs among individuals with respect to the identity of the nucleotide pair that occupies the site. For example, some DNA molecules may have a T–A base pair at a particular nucleotide site, whereas other DNA molecules in the same population may have a C–G base pair at the same site. This difference constitutes a SNP. The SNP defines two alleles for which there could be three genotypes among individuals in the population: homozygous with T–A at the corresponding site in both homologous chromosomes, homozygous with C–G at the corresponding site in both homologous chromosomes, or heterozygous with T–A in one chromosome and C–G in the homologous chromosome. In the human genome, any two randomly chosen DNA molecules are likely to differ at one SNP site

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

about every 1000 bp in noncoding DNA and at one SNP site about every 3000 bp in protein-coding DNA. Note, in the discussion of a SNP, the stipulation is that DNA molecules must differ at the nucleotide site “often.” This provision excludes rare genetic variations of the sort found in less than 1 percent of the DNA molecules in a population. Extremely rare genetic variants are not generally as useful in genetic analysis as the more common variants.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

SNPs are the most common form of genetic differences among people. Approximately 10 million SNPs have been identified that are relatively common in the human population, and 300,000– 600,000 of these are typically used in a search for SNPs that might be associated with complex diseases such as diabetes or high blood pressure (see the Genetic Linkage and Chromosome Mapping and The Genetic Basis of Complex Traits chapters). Identifying the particular nucleotide present at each of a million SNPs is made possible through the use of DNA microarrays containing millions of infinitesimal spots on a glass slide about the size of a postage stamp. Each tiny spot contains a unique DNA oligonucleotide sequence present in millions of copies synthesized by microchemistry when the microarray is manufactured. Each oligonucleotide sequence is designed to hybridize specifically with small fragments of genomic DNA that include one or the other of the nucleotide pairs present in a SNP. The microarrays also include numerous controls for each hybridization. The controls consist of oligonucleotides containing deliberate mismatches that are intended to guard against being misled by particular nucleotide sequences that are particularly “sticky” and hybridize too readily with genomic fragments and other sequences that form structures that hybridize poorly or not at all. Such microarrays, sometimes called SNP

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

chips, enable the SNP genotype of an individual to be determined with nearly 100 percent accuracy. The principles behind oligonucleotide hybridization are illustrated in FIGURE 2.27. In the figure, the length of each oligonucleotide is seven nucleotides; in practice, this is too short, as typical SNP chips consist of oligonucleotides at least 25 nucleotides in length. Part A shows the two types of DNA duplexes that might form a SNP. In this example, some chromosomes carry a DNA molecule

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

with a T–A base pair at the position shown in red, whereas the DNA molecular in other chromosomes has a C–G base pair at the corresponding position. Short fragments of genomic DNA are labeled with a fluorescent tag, and then the single strands are hybridized with a SNP chip containing the complementary oligonucleotides as well as the numerous controls. The duplex containing the T–A will hybridize only with the two oligonucleotides on the left, and that containing the C–G will hybridize only with the two oligonucleotides on the right.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.27 (A) Oligonucleotides attached to a glass slide in a SNP chip can be used to identify duplex DNA molecules containing alternative base pairs for a SNP—in this example, a T–A base pair versus a C–G base pair. (B) The SNP genotype of an individual can be determined by hybridization because DNA samples from genotypes that are homozygous TA/TA, homozygous CG/CG, or heterozygous TA/CG all give different patterns of fluorescence.

After the hybridization takes place, the SNP chip is examined with fluorescence microscopy to detect the spots that fluoresce due to the tag on the genomic DNA. The possible patterns are shown in part B. Genomic DNA from an individual whose chromosomes contain two copies of the TA form of the duplex (homozygous TA/TA) will cause the two leftmost spots to fluoresce, but the two rightmost spots will remain unlabeled. Similarly, genomic DNA from

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

a homozygous CG/CG individual will cause the two rightmost spots to fluoresce, but not the two spots on the left. Finally, genomic DNA from a heterozygous TA/CG individual will cause fluorescence of all four spots, because the TA duplex labels the two leftmost spots and the CG duplex labels the two rightmost spots. Use of SNP chips or other available technologies for highthroughput genotyping of millions of SNPs in thousands of individuals allows genetic risk factors for disease to be identified. A typical study compares the genotypes of patients with particular diseases with the genotypes of healthy people matched with the patients on such factors as sex, age, and ethnic group. Comparing the SNP genotypes among these groups often reveals which SNPs in the genome mark the location of the genetic risk factors, as will be explained in greater detail in Section 2.7.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Tandem Repeat Polymorphisms An important type of DNA polymorphism results from differences in the number of copies of a DNA sequence that may be repeated many times in tandem at a particular site in a chromosome. Any particular chromosome may have any number of copies of the tandem repeat, typically ranging from ten to a few hundred. FIGURE 2.28 illustrates DNA molecules that differ in the number of tandem repeats. In this case, the number of repeats varies from 1 to 10.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 2.28 A genetic polymorphism in which the alleles in a population differ in the number of copies of a DNA sequence (typically 2–60 bp) that is repeated in tandem along the chromosome. This example shows alleles in which the repeat number varies from 1 to 10. Amplification using primers flanking the repeat yields a unique fragment length for each allele.

THE CUTTING EDGE: High-Throughput SNP Genotyping

N

aturally occurring enzymes are often used for critical steps in molecular genetic analysis. Using restriction enzymes and

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

DNA polymerase for DNA marker analysis and PCR are two examples, and the tremendous diversity of species (especially microbial) on Earth continues to be a source of new tools. SNP detection is a good example. The basic SNP chip approach, as shown in Figure 2.27, depends on the availability of sufficient genomic DNA to be analyzed; it also requires accurate hybridization of genomic DNA fragments to oligonucleotides on

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the chip. Both of these limitations have been addressed by the latest techniques. To increase the quantity of genomic DNA, the technique of PCRfree whole-genome amplification has been developed. PCR by itself has two problems. First, it requires thermal cycling for the denaturation, renaturation, and elongation steps. Second, the Taq polymerase (the one most commonly used in PCR) is prone to errors: An incorrect base is inserted every 800 bases or so. To get around these problems, scientists have developed a technique for amplification of whole genomes called multiple displacement amplification. As illustrated in FIGURE A, random six-base sequences (hexamers) are used as primers. The DNA polymerase comes from the bacteriophage φ29, an enzyme that replicates DNA with high fidelity. Furthermore, when it encounters the 5′ end of a strand that is base-paired to its template, the DNA polymerase displaces that strand; the resulting single strand can then hybridize with another of the hexamer primer sequences present in the reaction, thereby serving as the template for further replication. By this means, total genomic DNA can be amplified from a single cell, and the process occurs without thermal cycling (no denaturation and renaturation is necessary).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE A

Once this phase is complete, the amplified DNA is sheared into small fragments, denatured, and hybridized to oligonucleotides (FIGURE B). Multiple copies of different oligonucleotides have been attached to microbeads, with their 3′ ends free, and those beads have been embedded in a solid matrix. However, the SNP to be assayed is not part of the sequence included in the oligonucleotide; rather, the SNP is the next base in the sequence after the oligonucleotide's 3′ end. DNA polymerase is added, along with modified nucleoside triphosphates that can be detected fluorescently, under conditions where one base will be added to the oligonucleotide probe. The identity of that base depends on the SNP allele in the hybridized genomic DNA, and it can be determined by scanning the fluorescence on the chip. In the case illustrated in Figure B the individual being assayed is homozygous for the A allele of SNP 1, heterozygous for the A and G alleles of SNP 2, and homozygous for the G allele of SNP 3.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE B

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

These kinds of cutting-edge methods make SNP genotyping fast and efficient with even only a very small amount of starting material (such as the saliva samples used by the personal genomics company 23andMe), and they can generate enormous amounts of data. In fact, as many as 48 samples can be genotyped for 300,000 or more SNPs in as little as three days. These methods do require some sophisticated instrumentation (much of it can be done robotically), but they remain fundamentally grounded in the biological processes of base pairing and DNA replication.

When any of the DNA molecules is cleaved with a restriction endonuclease that cleaves at sites flanking the tandem repeat, the size of the resulting restriction fragment is determined by the number of repeats that it contains. A molecule of duplex DNA containing the repeats can also be amplified by means of the polymerase chain reaction, using primers flanking the tandem repeat. Whether obtained by endonuclease digestion or PCR, the resulting DNA fragments increase in size according to the number of repeats they contain. Therefore, as shown at the right in Figure

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.28, two DNA molecules that differ in the number of copies of the tandem repeat can be distinguished because each will produce a different-sized DNA fragment that can be separated by means of amplification of that fragment. The acronym SSR stands for simple sequence repeat (these sequences are also known as microsatellite loci). An example of an SSR is the repeating sequence 5′-…ACACACAC…-3′. SSRs are important in genetic analysis for two reasons: ■ SSRs are abundant in the genomes of most eukaryotes. ■ SSRs are often highly polymorphic.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

To illustrate their abundance in the human genome, the most common SSRs are tabulated in TABLE 2.3. There are many more dinucleotide repeats than trinucleotide repeats, but there are great differences in abundance within each class. The most common dinucleotide repeat is 5′-…ACACACAC…-3′, which is present at more than 80,000 locations throughout the human genome. On average, the human genome has one SSR in every 2 kb of human DNA, or about 1.5 million SSRs altogether.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

TABLE 2.3 Some simple sequence repeats in the human genome SSR repeat unit

Number of SSRs in the human genome

5′-AC-3′

80,330

5′-AT-3′

56,260

5′-AG-3′

23,780

5′-GC-3′

290

5′-AAT-3′

11,890

5′-AAC-3′

7,540

5′-AGG-3′

4,350

5′-AAG-3′

4,060

5′-ATG-3′

2,030

5′-CGG-3′

1,740

5′-ACC-3′

1,160

5′-AGC-3′

870

5′-ACT-3′

580

Data from International Human Genome Sequencing Consortium, Nature. 2001; 409: 860–921.

Not only do SSRs tend to be polymorphic, but most polymorphisms tend to have a large number of alleles. Each “allele” corresponds to a DNA molecule with a different number of copies of the repeating unit, as illustrated in Figure 2.28. In other words, a polymorphism in the number of tandem repeats usually has multiple alleles in the population. Even with multiple alleles, however, any particular

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

chromosome must carry only one of the alleles defined by the number of tandem repeats, and any individual genotype can carry at most two different alleles. Nevertheless, a large number of alleles implies an even larger number of genotypes. For example, even with only 10 alleles in a population, there could be 10 different homozygous genotypes and 45 different heterozygous genotypes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

More generally, with n alleles there are a total of n(n + 1)/2 possible genotypes, of which n are homozygous and n(n − 1)/2 are heterozygous. Consequently, if an SSR has a relatively large number of alleles in a population, but no one allele is exceptionally common, then each of the many genotypes in the population will have a relatively low frequency. If the genotypes at 6–8 highly polymorphic loci are considered simultaneously, then each possible multiple-locus genotype is exceedingly rare. The very low frequency of any multiple-locus genotype is what gives tandemrepeat polymorphisms their utility in DNA typing (sometimes called DNA fingerprinting) for individual identification and for assessing the degree of genetic relatedness between individuals. In forensic DNA profiling, for example, genotypes are determined for 13 SSR loci that are distributed throughout the human genome; the probability that two individuals (other than unrelated twins) will have identical genotypes at all 13 loci is less than one in a billion. The use of DNA typing in criminal investigation is exemplified in Problem 2.24 and Challenge Problem 2 at the end of this chapter and discussed further in the Genes in Populations chapter. Technological advances have done much to facilitate DNA typing. To this point, we have considered only electrophoretic analyses performed manually. In fact, SSR genotype is routinely performed using automated systems, in which multiple loci can be genotyped in a single reaction. In this process (FIGURE 2.29), PCR is performed using multiple primers simultaneously, and the primers

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

are labeled with chemical tags that can be detected fluorescently. The resulting products are then separated electrophoretically in a gel contained in a capillary tube. As the molecules emerge from the tube, they are detected by a photodetector and their sizes recorded. Figure 2.29B shows the results of such an analysis. In this case, 24 SSR loci were genotyped from a starting sample of 1.2 picograms of human genomic DNA—comparable to the amount of DNA that can be obtained from a single cheek swab.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.



Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE 2.29 SSR genotyping with automated multiplexed PCR. (A) Schematic diagram showing four loci being amplified, each with primers carrying different fluorescent labels (blue, green, purple, and red) and producing different-sized fragments. (B) An example of SSR genotyping results. In this case, 24 different sets of SSR locus primers were used in combination with four different fluorescent markers. Shaded boxes indicate the names of the loci. If an individual is homozygous (in this case, D2S1338 and D12S391), a single peak is observed. If an individual is heterozygous, two peaks are seen. (B) Courtesy of Promega Corporation. Retrieved from

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

http://www.promega.com/~/media/images/resources/figures/1090010999/10911ma_800px.jpg?h=650&la=en&w=800.

Copy-Number Variation In addition to the small-scale variation in copy number represented by such genomic features as tandem repeats of short sequences (SSRs), a substantial portion of the human genome can be duplicated or deleted in much larger but still submicroscopic chunks

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

ranging from 1 kb to 1 Mb (Mb stands for megabase pairs, or 1 million base pairs). This type of variation is known as copynumber variation (CNV). The extra or missing copies of the genome in CNVs can be detected by means of hybridization with oligonucleotides in DNA microarrays. Because each spot on the microarray consists of millions of identical copies of a particular oligonucleotide sequence, the number of these sequences that undergo hybridization depends on how many copies of the complementary sequence are present in genomic DNA. A typical

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

region is present in two copies (one inherited from the mother and the other from the father). If an individual has an extra copy of the region, the ratio of hybridization and therefore fluorescence intensity will be 3 : 2; by comparison, if an individual has a missing copy, the ratio of hybridization and therefore fluorescence intensity will be 1 : 2. These differences can readily be detected with DNA microarrays. Moreover, because CNVs are relatively large, a microarray typically will include many different oligonucleotides that are complementary to sequences at intervals across the CNV; hence, the CNV will result in an increase or decrease in signal intensity of all the oligonucleotides included in the CNV. Current SNP chips also include about 1 million oligonucleotide probes designed to detect known CNVs. CNVs, by definition, exceed 1 kb in size, but many are much larger. In one study of approximately 300 individuals with ancestry tracing to Africa, Europe, or Asia, approximately 1500 CNVs were discovered by hybridization with microarrays. These averaged 200–300 kb in length. In the aggregate, the CNVs included 300– 450 million base pairs, or 10–15 percent of the nucleotides in the entire genome. Many of the CNVs were located in regions near known mutant genes associated with hereditary diseases. CNVs in alpha and beta hemoglobin genes are known to be associated with resistance to malaria, and CNVs in the HIV-1 receptor gene CCL3

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

are associated with resistance to AIDS. CNVs have been reported

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

to be risk factors for complex diseases such as Alzheimer's disease, autism, and schizophrenia.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.7 Applications of DNA Markers Why are geneticists interested in DNA markers and DNA polymorphisms? Their interest can be justified on any number of grounds. In this section, we consider the reasons most often cited.

Genetic Markers, Genetic Mapping, and “Disease Genes”

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Perhaps the key goal in studying DNA polymorphisms in human genetics is to identify the chromosomal location of genes having mutant alleles associated with hereditary diseases. In the context of disorders caused by the interaction of multiple genetic and environmental factors, such as heart disease, cancer, diabetes, depression, and so forth, it is important to think of a harmful allele as a risk factor for the disease that increases the probability of occurrence of the disease, rather than as the sole causative agent. This point needs to be emphasized, especially because genetic risk factors are often called disease genes. For example, a major “disease gene” for breast cancer in women is the gene BRCA1. For women who carry a mutant allele of BRCA1, the lifetime risk of breast cancer is about 60 percent. By comparison, among women who are not carriers, the lifetime risk of breast cancer is about 12 percent. Hence, many women without the genetic risk factor do develop breast cancer. Indeed, BRCA1 mutations are found in only 16 percent of affected women who have a family history of breast cancer. The importance of a genetic risk factor can be expressed quantitatively as the relative risk, which equals the ratio of affected persons to nonaffected persons among those individuals who carry the risk factor divided by the ratio of affected persons to

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

nonaffected persons among those individuals who do not carry it. In the case of BRCA1, for example, the relative risk is (0.60/0.40 = 1.5) divided by (0.12/0.88 = 0.136), which equals 11. The utility of DNA polymorphisms in locating and identifying disease

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

genes results from genetic linkage, or the tendency for genes that are sufficiently close together in a chromosome to be inherited together. To see how this works, refer back to Figure 2.1. In that hypothetical example, the DNA marker is not actually in the PAH gene, but rather one marker allele is associated with the disease allele as a result of genetic linkage. Genetic linkage is discussed in detail in the Genetic Linkage and Chromosome Mapping chapter, but the key concepts are summarized in FIGURE 2.30, which shows the location of many DNA polymorphisms along a chromosome that also carries a genetic risk factor denoted D (for disease gene). Each DNA polymorphism serves as a genetic marker for its own location in the chromosome. The importance of genetic linkage is that alleles of DNA markers that are sufficiently close to the disease gene will tend to be inherited together with the disease gene in pedigrees—and the closer the markers, the stronger this association. Hence, the initial approach to the identification of a disease gene is to find DNA markers that are genetically linked with the disease gene so as to identify its chromosomal location, a procedure known as genetic mapping. Once the chromosomal position is known, other methods can be used to pinpoint the disease gene itself and to study its functions.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

FIGURE 2.30 Concepts in genetic localization of genetic risk

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

factors for disease. Polymorphic DNA markers (indicated by the vertical lines) that are close to a genetic risk factor (D) in the chromosome tend to be inherited together with the disease itself. The genomic location of the risk factor is determined by examining the known genomic locations of the DNA polymorphisms that are linked with it.

If genetic linkage seems a roundabout way to identify disease genes, consider the alternative. The human genome contains approximately 30,000 genes. If genetic linkage did not exist, then we would have to examine 30,000 DNA polymorphisms, one in each gene, to identify a disease gene. But the human genome has only 23 pairs of chromosomes, and because of genetic linkage and the power of genetic mapping, it actually requires only a few hundred DNA polymorphisms to identify the chromosome and approximate location of a genetic risk factor.

Other Uses for DNA Markers DNA polymorphisms are widely used in all aspects of modern genetics because they provide a large number of easily accessed genetic markers for genetic mapping and other purposes. Some other uses of DNA polymorphisms are discussed in this section.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Individual identification. We have already mentioned that DNA polymorphisms have application as a means of DNA typing (DNA fingerprinting) to identify different individuals in a population. DNA typing in other organisms is used to determine individual animals in endangered species and to identify the degree of genetic relatedness among individual organisms that live in packs or herds. For example, DNA typing in wild horses has shown that the wild stallion in charge of a harem of mares actually sires fewer than one-third of the foals.

Epidemiology and food safety science. DNA typing has important applications in tracking the spread of viral and bacterial epidemic diseases, as well as in identifying the source of contamination in contaminated foods.

Human population history. DNA polymorphisms provide important information that enables anthropologists to reconstruct the evolutionary origin, global expansion, and diversification of the human population. We will consider this issue in depth in the Molecular and Human Evolutionary Genetics chapter.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Improvement of domesticated plants and animals. Plant and animal breeders have turned to DNA polymorphisms as genetic markers in pedigree studies to identify, by genetic mapping, genes that are associated with favorable traits. These genes may then be incorporated into currently used varieties of plants and breeds of animals.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

History of domestication. Plant and animal breeders also study genetic polymorphisms to identify the wild ancestors of cultivated plants and domesticated animals, as well as to infer the practices of artificial selection that led to genetic changes in these species during domestication.

DNA polymorphisms as ecological indicators. DNA polymorphisms are being evaluated as biological indicators of genetic diversity in key indicator species present in biological communities exposed to chemical, biological, or physical stress. They are also used to monitor genetic diversity in endangered species and species bred in captivity.

Evolutionary genetics. DNA polymorphisms are studied in an effort to describe the patterns in which different types of genetic variation occur throughout the genome, to infer the evolutionary mechanisms by which genetic variation is maintained, and to illuminate the processes by which genetic polymorphisms within species become transformed into genetic differences between species.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Population studies. Population geneticists employ DNA polymorphisms to assess the level of genetic variation in diverse populations of organisms that differ in genetic organization (prokaryotes, eukaryotes, organelles), population size, breeding structure, or life-history characteristics. They use genetic polymorphisms within subpopulations of a species as indicators of population history, patterns of migration, and so forth.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Evolutionary relationships among species. Differences in DNA sequences between species serve as the basis of molecular phylogenetics, in which the sequences are analyzed to determine the ancestral history (phylogeny) of the species and to trace the origin of morphological, behavioral, and other types of adaptations that have arisen in the course of evolution.

SUMMING UP ■ Analysis of RFLPs was an important tool for initial mapping of the human genome. ■ Single-nucleotide polymorphisms occur, on average, in one of every 1000 bases, so they provide a wealth of potential genetic markers. ■ DNA microarray technology is used to simultaneously genotype individuals at hundreds of thousands of SNP loci. ■ Simple sequence repeats are also ubiquitous in the human genome and are widely used in DNA forensics.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Use of DNA markers for genetic analysis has been applied to areas ranging from epidemiology to evolutionary genetics.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

CHAPTER SUMMARY ■ A DNA strand is a polymer of A, T, G, and C deoxyribonucleotides joined in the 3′-to-5′ direction by phosphodiester bonds. ■ The two DNA strands in a duplex are held together by hydrogen bonding between the A–T and G–C base pairs and by base stacking of the paired bases. ■ Each type of restriction endonuclease enzyme cleaves doublestranded DNA at a particular sequence of bases that is usually four or six nucleotides in length. ■ The DNA fragments produced by a restriction enzyme can be separated by electrophoresis, isolated, sequenced, and manipulated in other ways. ■ Separated strands of DNA or RNA that are complementary in nucleotide sequence can come together (hybridize) spontaneously to form duplexes. ■ DNA replication takes place only by elongation of the growing strand in the 5′-to-3′ direction through the addition of successive nucleotides to the 3′ end.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ In the polymerase chain reaction, short oligonucleotide primers are used in successive cycles of DNA replication to amplify selectively a particular region of a DNA duplex. ■ Genetic markers in DNA provide a large number of easily accessed sites in the genome that can be used to identify the chromosomal locations of disease genes, for DNA typing in individual identification, for the genetic improvement of cultivated plants and domesticated animals, and for many other applications.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

REVIEW THE BASICS ■ Define and give an example of each of the following genetic terms: locus, allele, genotype, heterozygous, homozygous, phenotype. ■ What is a DNA marker? Explain how harmless DNA markers can serve as aids in identifying disease genes through genetic mapping. ■ Which four bases are commonly found in the nucleotides in DNA? Which form base pairs? ■ Which chemical groups are present at the extreme 3′ and 5′ ends of a single polynucleotide strand? ■ What does it mean to say that a single strand of DNA strand has a polarity? What does it mean to say that the DNA strands in a duplex molecule are antiparallel? ■ What are restriction enzymes, and why are they important in the study of particular DNA fragments? What does it mean to say that most restriction sites are palindromes? ■ Describe how a Southern blot is carried out. What is it used for? What is the role of the probe?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ How does the polymerase chain reaction work? What is it used for? Which information about the target sequence must be known in advance? What is the role of the oligonucleotide primers? ■ What is a DNA microarray? How is one used for SNP genotyping? What are the possible sources of error in the process? ■ What is a DNA marker? Explain how harmless DNA markers can serve as aids in identifying disease genes through genetic mapping.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

GUIDE TO PROBLEM SOLVING PROBLEM 1 A geneticist plans to use the polymerase chain reaction (PCR) to amplify part of the DNA sequence shown below, using oligonucleotide primers that hybridize in the regions marked in red. (These are illustrative only and too short to be used in practice.) Specify the sequence of the primers that should be used, including the polarity, and deduce the sequence of the amplified DNA fragment.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

ANSWER The primers should be able to pair with the chosen primer sites and must be oriented with their 3′ ends facing each other. Thus the “forward” primer (the one that is elongated in the left-to-right direction) should have the sequence 5′-ACGAT-3′, and the “reverse” primer (the one that is elongated in the right-to-left direction) should have the sequence 3′-ACTAG-5′. Because the convention for writing nucleic acid sequences is to put 5′ end on the left, the reverse primer is 5′-GATCA-3′. PROBLEM 2 The genetic material of the bacteriophage M13 is a single-stranded DNA molecule of 6407 nucleotides, whose complete sequence can be found in publicly accessible databases such as GenBank. Upon M13 DNA being introduced into a bacterial cell, a complementary DNA strand is synthesized by bacterial enzymes, resulting in a double-stranded replicative form of the phage genome. How would you determine whether both strands of the replicative form are transcribed within an infected cell? ANSWER To determine whether both DNA strands are templates for RNA synthesis, one could isolate the replicative form of the phage DNA, separate the two strands, and test each strand

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

separately for the ability to hybridize with mRNA found in M13infected cells. Alternatively (and less definitively), one could examine the genome sequence of M13 in GenBank and search both DNA strands for open reading frames that could code for proteins. As it happens, only one strand of M13 contains such open reading frames, and hence only one DNA strand is likely to be transcribed. PROBLEM 3 A plasmid of 6200 base pairs includes at least three complete copies of a viral gene arranged in tandem. Cleavage of the plasmid with the restriction enzyme HindIII results in fragments of 900, 1300, and 4000 bp. The tandem copies of the gene are contained within only one of the HindIII fragments. The gene encodes a protein of 405 amino acids. Which HindIII fragment is likely to contain the gene copies?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

ANSWER To encode a protein of 405 amino acids requires at least 1215 bp. Three tandem copies of the gene would therefore require 3645 bp, and four copies 4860 bp. The smallest HindIII fragment is not long enough to contain even one copy of the gene. The 1300bp fragment could contain one copy but not two or more. Only the largest fragment could include three copies of the gene, and it could not include more than three copies. PROBLEM 4 Factor V is a protein that is involved in the blood clotting cascade. Men of European ancestry who are heterozygous for a particular rare allele of the F5 gene that encodes it have a 52 percent chance of developing venous thromboembolism, a blood clot forming in a vein in the leg. In the population as a whole, 12 percent of all similar men develop this problem. What is the risk factor associated with this particular genotype?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

ANSWER If 52 percent of men with the risk-associated genotype develop the problem, then 100 percent − 52 percent, or 48 percent, do not. Thus, we can calculate the ratio of men with this genotype who do and do not develop the condition as 52/48, or 1.08. Next, we need to be able to characterize the risk of venous thromboembolism occurring regardless of genotype. We do the same calculation for the population as a whole, where 12 percent of men develop the condition and 88 percent do not. The corresponding ratio is thus 12/88, or 0.136. Finally, we can determine the increased risk associated with F5 genotype as

Thus, the risk of developing venous thromboembolism for a European man heterozygous for the risk-associated F5 allele is 7.9 times higher than it is for the European male population as a whole.

ANALYSIS AND APPLICATIONS

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.1 Which chemical groups are present at the 3′ and 5′ ends of a single-polynucleotide molecule? 2.2 In the deoxyribonucleotide shown below, which carbon atom carries the phosphate group and which carries the 3′ hydroxyl group?

2.3 Many restriction enzymes produce restriction fragments that have “sticky ends.” What does this mean?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.4 Which of the following sequences are palindromes and which are not? Explain your answer. (a) 5′-CCGG-3′ (b) 5′-TTTT-3′ (c) 5′-GCTAGC-3′ (d) 5′-CCGCTC-3′ (e) 5′-AAGGTT-3′

2.5 The list below gives half of each of a set of palindromic restriction sites. Replace the Ns to complete the sequence of each restriction site. (a) 5′-AGNN-3′ (b) 5′-ATGNNN-3′ (c) 5′-ATTNNN-3′ (d) 5′-NNNAGC-3′

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.6 Apart from nucleotide sequence, what is different about the ends of restriction fragments produced by the following restriction enzymes? (The downward arrow represents the site of cleavage in each strand.) (a) ScaI (5′-AGT↓ACT-3′) (b) NheI (5′-G↓CTAGC-3′) (c) CfoI (5′-GCG↓C-3′)

2.7 A solution contains double-stranded DNA fragments of size 4 kb, 8 kb, 10 kb, and 13 kb that are separated in an electrophoresis gel. In the accompanying diagram of the gel, match the fragments sizes with the correct bands.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.8 The linear DNA fragment shown here has cleavage sites for AatII (A) and XhoI (X). In the accompanying diagram of an electrophoresis gel, indicate the positions at which bands would be found after digestion with: (a) AatII alone. (b) XhoI alone. (c) AatII and XhoI together.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The dashed lines on the right indicate the positions to which bands of 1–12 kb would migrate.

2.9 The circular DNA molecule shown here has cleavage sites for AatII and XhoI. In the accompanying

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

diagram of an electrophoresis gel, indicate the positions at which bands would be found after digestion with: (a) AatII alone. (b) XhoI alone. (c) AatII and XhoI together.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The dashed lines indicate the positions to which bands of 1– 12 kb would migrate.

2.10 Consider the accompanying diagram of a region of duplex DNA, in which the Bs represent bases in Watson–Crick pairs. Specify as precisely as possible the identity of: (a) B5, assuming that B1 = A. (b) B6, assuming that B2 = C.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

(c) B7, assuming that B3 = any purine. (d) B8, assuming that B4 = A or T.

2.11 In the precursor nucleotides of the DNA duplex diagrammed in Problem 2.10, with which base was each of the phosphate groups 1–4 associated with prior to its incorporation into the polynucleotide strand?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.12 In a random sequence consisting of equal proportions of all four nucleotides, what is the probability that a particular short sequence of nucleotides matches a restriction site for: (a) A restriction enzyme with a 4-base cleavage site? (b) A restriction enzyme with a 6-base cleavage site? (c) A restriction enzyme with an 8-base cleavage site?

2.13 In a random sequence consisting of equal proportions of all four nucleotides, what is the average distance between restriction sites for: (a) A restriction enzyme with a 4-base cleavage site?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

(b) A restriction enzyme with a 6-base cleavage site? (c) A restriction enzyme with an 8-base cleavage site?

2.14 If Escherichia coli DNA were essentially a random sequence of 4.6 × 106 bp with equal proportions of all four nucleotides (this is an oversimplification), approximately how many restriction fragments would be expected from cleavage with: (a) A “4-cutter” restriction enzyme? (b) A “6-cutter” restriction enzyme? (c) An “8-cutter” restriction enzyme?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.15 A circular DNA molecule is cleaved with AfeI, NheI, or both restriction enzymes together. The accompanying diagram shows the resulting electrophoresis gel, with the band sizes indicated. Draw a diagram of the circular DNA showing the relative positions of the AfeI and NheI sites.

2.16 In the diagrams of DNA fragments shown here, the tick marks indicate the positions of restriction sites for a particular restriction enzyme. A mixture of the two types of molecules is digested and analyzed with a Southern blot using either probe A or probe B, which hybridize to the fragments at the positions shown by the rectangles. In the accompanying gel

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

diagram, indicate the bands that would result from each of these probes. (The scale on the right shows the expected positions of fragments from 1–12 kb.)

2.17 In the accompanying diagram, the tick marks indicate the positions of restriction sites in two alternative DNA fragments that can be present at a locus in a human chromosome. An RFLP analysis is carried out, using probe DNA that hybridizes with the fragments at the position shown by the rectangle. With respect to this RFLP, how many genotypes are possible? (Use the symbol A1 to refer to the allele that yields the upper DNA fragment, and A2 to refer to the allele that yields the lower DNA fragment.) In the accompanying gel diagram, indicate the genotypes across the top and the phenotype (band position or positions) expected of each genotype. (The scale on the right shows the expected positions of fragments from 1 to 12 kb.)

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.18 If pentamers were long enough to serve as specific oligonucleotide primers for PCR (in practice, they are too short), which DNA fragment would be amplified using the “forward” primer 5′-AATGC-3′ and the “reverse” primer 3′-GCATG-5′ acting on the doublestranded DNA molecule shown here?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.19 Would the primer pairs 3′-AATGC-5′ and 5′-GCATG3′ amplify the same fragment described in Problem 2.18? Explain your answer. 2.20 Suppose that a fragment of human DNA of length 3 kb is to be amplified by PCR. The total genome size is 3 × 106 kb, which equals 3 × 109 base pairs. (a) Prior to amplification, what fraction of the total DNA does the target sequence constitute? (b) What fraction does it constitute after 10 cycles of PCR? (c) After 20 cycles of PCR? (d) After 30 cycles of PCR?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

2.21 The two allelic fragments of DNA shown below are amplified by PCR, and the products digested with both PstI and Hind3. (a) What will the total sizes of the two PCR products be?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(b) Diagram the gel pattern you would expect to see from the doubly digested DNA.

2.22 Below are diagrammed six allelic DNA fragments, showing a number of SNP polymorphisms. Three of these fragments also carry an allele of a gene associated with an inherited disorder (indicated by D). As a clinical geneticist, you would like to use these SNP markers to determine whether an individual is at increased risk of getting the disease. Which of these SNPs would assist you in that determination? Which would not? Justify your answers.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.23 Psoriasis is a skin disease that has been shown to have a significant genetic component, although, as with breast cancer, multiple genes contribute to its development. An allele of one such gene, known as PSORS1, has been located on chromosome 6, as has a SNP marker (rs10484554, a C/T polymorphism); the T allele of this SNP is associated with the risk allele of PSORS1. In a population study, 22.4 percent of heterozygous (genotype CT) individuals contracted psoriasis, while in a control population of individuals with genotype CC, the incidence of psoriasis was 11.4 percent. What is the relative risk associated with this genotype? 2.24 A cigarette butt found at the scene of a robbery is found to have a sufficient number of epithelial cells stuck to the paper that the DNA can be extracted and analyzed by DNA typing. Shown here are the results of typing for three probes (locus 1–locus 3) of the evidence (X) and cells from seven suspects (A–G). Which of the suspects can be excluded? Which cannot be excluded? Can you identify the perpetrator? Explain your reasoning.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.25 A woman is uncertain which of two men is the father of her child. DNA typing is carried out on blood from the child (C), the mother (M), and the two males (A and B), using probes for a highly polymorphic DNA marker on two different chromosomes (“locus 1” and “locus 2”). The result is shown in the accompanying diagram. Does either or both of the tested loci rule out any of the males as being the possible father? Explain your reasoning.

2.26 Snake venom diesterase cleaves the chemical bonds shown in red in the accompanying diagram,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

leaving mononucleotides that are phosphorylated in the 3′ position. If the phosphates numbered 2 and 4 are radioactive, which mononucleotides will be radioactive after cleavage with snake venom diesterase?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.27 A DNA sample collected from the scene of a crime is believed to be from the perpetrator. It is genotyped for four SSR loci. Four suspects are genotyped as well. All five genotypes are shown below as fluorometric traces, with the colors corresponding to different loci. (a) Which individuals can be excluded as suspects and why? (b) Which of the individuals could be the perpetrator? Do you think the evidence is conclusive? Why or why not?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 The genome of Drosophila melanogaster is 180 × 106 bp, and a fragment of size 1.8 kb is to be amplified by PCR. How many cycles of PCR are necessary for the amplified target sequence to constitute at least 99 percent of the total DNA?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

CHALLENGE PROBLEM 2 The body of a young victim of murder is found in an advanced state of decomposition and cannot be identified. Police suspect the victim is one of five persons reported by their parents as missing. DNA typing is carried out on tissues from the victim (X) and on the five sets of parents (A–E), using probes for a highly polymorphic DNA marker on two different chromosomes (“locus 1” and “locus 2”). The result is shown in the accompanying diagram. How do you interpret the fact that genomic

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

DNA from each individual yields two bands? Can you identify the parents of the victim? Explain your reasoning.

CHALLENGE PROBLEM 3 The snake venom diesterase enzyme described in Problem 2.26 was originally used in a procedure called “nearest neighbor” analysis. In this procedure, a DNA strand is synthesized in the presence of all four nucleoside triphosphates, one of which carries a radioactive phosphate in the (innermost) position. Then the DNA is digested to completion with snake venom diesterase, and the resulting mononucleotides are separated and assayed for radioactivity. Examine the diagram in Problem 2.26 and explain:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

(a) How this procedure reveals the “nearest neighbors” of the radioactive nucleotide. (b) Whether the “nearest neighbor” is on the 5′ or the 3′ side of the labeled nucleotide.

FOR FURTHER READING Botstein, D., White, R. L., Skolnick, M., & Davis, R. W. (1980). Construction of a genetic linkage map in man using restriction fragment length polymorphisms. American Journal of Human Genetics, 32(3), 314–31. Retrieved from http://www.ncbi.nlm.nih.gov/pubmed/6247908 A classic paper describing how RFLP mapping could be used in human genetic analysis.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Kelly, T. J., & Smith, H. O. (1970). A restriction enzyme from Haemophilus influenzae: II. Base sequence of the recognition site. Journal of Molecular Biology, 51(2), 393–409. http://doi.org/10.1016/00222836(70)90150-6 One of the first reports of the recognition sequence of a restriction enzyme. Roewer, L. (2013). DNA fingerprinting in forensics: Past, present, future. Investigative Genetics, 4(1), 22. http://doiorg/10.1186/2041-2223-4-22

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

A recent overview of the status of forensic DNA testing. Southern, E. M. (1975). Detection of specific sequences among DNA fragments separated by gel electrophoresis. Journal of Molecular Biology, 98(3), 503–517. http://doi.org/10.1016/S00222836(75)80083-0 The original Southern blotting procedure. Steemers, F. J., Chang, W., Lee, G., Barker, D. L., Shen, R., & Gunderson, K. L. (2006). Wholegenome genotyping with the single-base extension assay. Nature Methods, 3(1), 31–33. http://doi.org/10.1038/nmeth842

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

A description of the basis for whole-genome SNP genotyping.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

© Molekuul/Science Photo Library/Getty Images.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

UNIT 2 Transmission Genetics Chapter 3 Mendelian Genetics: The Principles of Segregation and Assortment Chapter 4 The Chromosomal Basis of Inheritance Chapter 5 Genetic Linkage and Chromosome Mapping

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Chapter 6 Human Karyotypes and Chromosome Behavior Chapter 7 The Genetic Basis of Complex Traits

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Chapter 8 Genetics of Bacteria and their Viruses

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:17:58.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

© Biophoto Associates/Science Source.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

CHAPTER 3

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Mendelian Genetics: The Principles of Segregation and Assortment

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

CHAPTER OUTLINE 3.1 Morphological and Molecular Phenotypes 3.2 Segregation of a Single Gene 3.3 Segregation of Two or More Genes 3.4 Human Pedigree Analysis 3.5 Incomplete Dominance and Epistasis 3.6 Probability in Genetic Analysis 3.7 Conditional Probability and Pedigrees ROOTS OF DISCOVERY: What Did Gregor Mendel Think He Discovered? Gregor Mendel (1866) Experiments on Plant Hybrids ROOTS OF DISCOVERY: This Land Is Your Land The Huntington's Disease Collaborative Research Group (1993)

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

A Novel Gene Containing a Trinucleotide Repeat That Is Expanded and Unstable on Huntington's Disease Chromosomes

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

LEARNING OBJECTIVES & SCIENCE COMPETENCIES When you have learned the basics of transmission genetics, including segregation and independent assortment, and are able to combine these with elementary principles of probability, you will have gained critical science competencies enabling you to solve problems such as the following: ■ For one gene, given the genotypes or phenotype of parents, apply the principle of segregation to predict the possible types of progeny and their expected proportions. ■ For two or more genes, given the genotypes or phenotype of parents, use the principle of independent assortment to predict the possible types of progeny and their expected proportions.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ For any specified departures from the Mendelian assumptions regarding dominance or epistasis, predict how the expected proportions of progeny types will be modified in any particular cross. ■ Identify the inheritance patterns of simple Mendelian dominant or recessive traits in human pedigrees. ■ Make inferences about the genotypes of individuals in pedigrees based on their own phenotypes and those of their close relatives. n this chapter, we consider how genes are transmitted from parents to offspring and how this determines the distribution of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

I

genotypes and phenotypes among related individuals. The study

of the inheritance of traits constitutes transmission genetics. This subject is also called Mendelian genetics because the underlying principles were first deduced from experiments with garden peas (Pisum sativum) carried out in the years 1856–1863 by Gregor Mendel, a monk at the monastery of St. Thomas in the town of Brno (Brünn), in the Czech Republic. Mendel reported his experiments to a local natural history society, published the results and his interpretation in its scientific journal in 1866, and began

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

exchanging letters with one of the leading botanists of the time. His experiments were careful and exceptionally well documented, and his paper contains the first clear exposition of the statistical rules governing the transmission of genes from generation to generation. Nevertheless, Mendel's paper was largely ignored for 34 years until its significance was finally appreciated.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.1 Morphological and Molecular Phenotypes Until the advent of molecular genetics, geneticists dealt mainly with morphological traits, in which the differences between organisms can be expressed in terms of traits such as color, shape, or size. Mendel studied seven morphological traits contrasting in seed

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

shape, seed color, flower color, pod shape, and so forth (FIGURE 3.1). Perhaps the most widely known example of a contrasting Mendelian trait is round versus wrinkled seeds. As pea seeds dry, they lose water and shrink. Round seeds are round because they shrink uniformly; wrinkled seeds are wrinkled because they shrink irregularly. The wrinkled phenotype is due to the absence of a branched-chain form of starch known as amylopectin, which is not synthesized in wrinkled seeds owing to a defect in the enzyme starch-branching enzyme I (SBEI).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.1 The seven different traits in peas studied by Mendel. The phenotype shown at the far right is the dominant trait, which appears in the hybrid produced by crossing.

The nonmutant, or wild-type, allele of the gene for SBEI is designated W and the mutant allele w. Seeds that are heterozygous Ww have only half as much SBEI enzyme as wildtype homozygous WW seeds, but even half the normal amount of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

enzyme produces enough amylopectin for the heterozygous Ww seeds to shrink uniformly and remain phenotypically round. Hence, with respect to seed shape, genotypes WW and Ww have the same phenotype (round). The W allele is called the dominant allele and w is called the recessive allele. The molecular basis of the wrinkled mutation is that the SBEI gene has become interrupted by the insertion, into the gene, of a DNA sequence called a transposable element. Such DNA sequences

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

are capable of moving (transposition) from one location to another within a chromosome or between chromosomes. The molecular mechanism of transposition is discussed in the Mutation, Repair, and Recombination chapter, but for our present purposes it is necessary to know only that transposable elements are present in most genomes, especially the large genomes of eukaryotes, and that many spontaneous mutations result from the insertion of transposable elements into a gene. FIGURE 3.2 includes a diagram of the DNA structure of the wildtype W and mutant w alleles and shows the DNA insertion that interrupts the w allele. Highlighted are polymerase chain reaction (PCR) primer sites, present in both alleles, that flank the site of the insertion. The diagram in part C indicates the pattern of bands that would be expected if one were to amplify genomic DNA and separate the resulting products by electrophoresis. The product from the W allele would be smaller than that of the w allele because of the inserted DNA in the w allele, so it would migrate faster than the corresponding fragment from the w allele and move to a position closer to the bottom of the gel. Genomic DNA from homozygous WW would yield a single, faster-migrating band; that from homozygous ww a single, slower-migrating band; and that from heterozygous Ww two bands with the same electrophoretic mobilities as those observed in the homozygous genotypes. In

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Figure 3.2C, the band in the homozygous genotypes is shown as somewhat thicker than those from the heterozygous genotype, because the single band in each homozygous genotype comes from the two copies of whichever allele is homozygous. Thus, this band contains more DNA than the corresponding DNA in the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

heterozygous genotype, in which only one copy of each allele is present.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.





FIGURE 3.2 (A) W (round) is an allele of a gene that specifies the amino acid sequence of starch branching enzyme I (SBEI). (B) w (wrinkled) is an allele that encodes an inactive form of the enzyme because its DNA sequence is interrupted by the insertion of a transposable element. (C) At the level of the morphological phenotype, W is dominant to w: Genotypes WW and Ww have round seeds, whereas genotype ww has wrinkled seeds. The molecular difference between the alleles can be detected as a restriction fragment length polymorphism (RFLP) using the enzyme EcoRI and a probe that hybridizes at the site shown. At the molecular level, the alleles are codominant: DNA from each genotype yields a different molecular phenotype—a single band differing in size for homozygous WW and ww, and both bands for heterozygous Ww.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

IN THIS SMALL GARDEN plot adjacent to the monastery of St. Thomas, Gregor Mendel grew more than 33,500 pea plants over a period of eight years, including more than 6400 plants in one year alone. He received some help from two fellow monks who assisted in the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

experiments.

As illustrated in Figure 3.2C, the PCR analysis clearly distinguishes between the genotypes WW, Ww, and ww, because the heterozygous Ww genotype exhibits both of the bands observed in the homozygous genotypes. This situation is described by saying that the W and w alleles are codominant with respect to the molecular phenotype. However, as indicated by the seed shapes in Figure 3.2C, W is dominant over w with respect to morphological phenotype. In the discussion that follows, we use the PCR analysis of the W and w alleles to emphasize the importance of molecular phenotypes in modern genetics and to demonstrate experimentally the principles of genetic transmission.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

SUMMING UP ■ Mendel studied seven morphological traits in garden peas. ■ Dominant alleles are ones that result in identical phenotypes in both homozygotes and heterozygotes. ■ For each trait that Mendel studied, there were two alleles— one dominant and one recessive.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Use of molecular markers can allow scientists to identify all genotypes—homozygous or heterozygous. Such markers are thus codominant.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.2 Segregation of a Single Gene Mendel selected peas for his experiments for two primary reasons. First, he had access to varieties that differed in contrasting traits, such as round versus wrinkled seeds and yellow versus green seeds. Second, his earlier studies had indicated that peas usually reproduce by self-pollination, in which pollen produced in a flower is used to fertilize the eggs in the same flower. Producing hybrids by

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

cross-pollination (outcrossing) required painstaking surgery on immature flowers to reveal the receptive female structures, excise and discard the undeveloped male pollen-producing structures, and expose the female structures to mature pollen from a different plant. After the cross-fertilization, each dissected flower was enclosed in a fine mesh bag to prevent stray pollen from accessing the female structures inside. The relatively small space needed to grow each plant, and the relatively large number of progeny that could be obtained, gave Mendel the opportunity, as he says in his paper, to “determine the number of different forms in which hybrid progeny appear” and to “ascertain their numerical interrelationships.” The fact that garden peas are normally self-fertilizing means that in the absence of deliberate outcrossing, plants with contrasting traits are usually homozygous for alternative alleles of a gene affecting the trait; for example, plants with round seeds have genotype WW and those with wrinkled seeds have genotype ww. The homozygous genotypes are indicated experimentally by the observation that hereditary traits in each variety are truebreeding, which means that plants produce only progeny like themselves when allowed to self-pollinate normally. Outcrossing between plants that differ in one or more traits creates a hybrid. If the parents differ in one, two, or three traits, the hybrid is a monohybrid, dihybrid, or trihybrid, respectively. In keeping track of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

parents and their hybrid progeny, we say that the parents constitute the P1 generation and their hybrid progeny the F1 generation. Because garden peas are sexual organisms, each cross can be performed in two ways, depending on which phenotype is present in the female parent and which in the male parent. With round versus wrinkled, for example, the female parent can be round and the male wrinkled, or the reverse. These are called reciprocal crosses. Mendel was the first to demonstrate the following important principle: The outcome of a genetic cross does not depend on which trait is present in the male and which is present in the female; reciprocal crosses yield the same result.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

This principle is illustrated for round and wrinkled seeds in FIGURE 3.3. The gel icons show the PCR products that genomic DNA from each type of seed in these crosses would yield. The genotypes of the crosses and their progeny are as follows:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.3 Morphological and molecular phenotypes showing the equivalence of reciprocal crosses. In this example, the hybrid seeds are round and yield a PCR pattern with two bands, irrespective of the direction of the cross.

In both reciprocal crosses, the progeny have the morphological phenotype of round seeds, but as shown by the PCR patterns on the right, all progeny genotypes are actually heterozygous Ww and therefore different from either parent. The genetic equivalence of reciprocal crosses illustrated in Figure 3.3 is a principle that is quite general in its applicability, but there is an important exception, having to do with sex chromosomes, that is discussed in the chapter titled The Chromosomal Basis of Inheritance.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

In the following section, we examine a few of Mendel's original experiments in the context of molecular analysis to relate the morphological phenotypes and their ratios to the molecular phenotypes that would be expected.

Phenotypic Ratios in the F2 Generation Although the progeny of the crosses in Figure 3.3 have the dominant phenotype of round seeds, the PCR analysis shows that they are actually heterozygous. The w allele is hidden with respect to the morphological phenotype because w is recessive to W. Nevertheless, the wrinkled phenotype reappears in the next generation when the hybrid progeny are allowed to undergo selffertilization. For example, if the round F1 seeds from cross A in Figure 3.3 were grown into sexually mature plants and allowed to undergo self-fertilization, some of the resulting seeds would be round and others wrinkled. The progeny seeds produced by selffertilization of the F1 generation constitute the F2 generation. When Mendel carried out this experiment, in the F2 generation he

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

observed the results shown in the following diagram:

Note that the ratio 5474 : 1850 is approximately 3 : 1.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

A 3 : 1 ratio of dominant to recessive forms in the F2 progeny is characteristic of simple Mendelian inheritance. Mendel's data demonstrating this point are shown in TABLE 3.1. Note that the first two traits in the table (round versus wrinkled seeds and yellow versus green seeds) have many more observations than any of the others. The reason is that these traits can be classified directly in the seeds, whereas the others can be classified only in the mature plants, and Mendel could analyze many more seeds than he could mature plants. The principal observations from the data in Table 3.1 are as follows: ■ The F1 hybrids express only the dominant trait (because the F1 progeny are heterozygous—for example, Ww). ■ In the F2 generation, plants with either the dominant trait or the recessive trait are present (which means that some F2 progeny are homozygous—for example, ww). ■ In the F2 generation, there are approximately three times as

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

many plants with the dominant phenotype as plants with the recessive phenotype.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

TABLE 3.1 Results of Mendel's monohybrid experiments Parental traits

F1 trait

Number of F2 progeny

F2 ratio

Round × wrinkled (seeds)

Round

5474 round, 1850 wrinkled

2.96 : 1

Yellow × green (seeds)

Yellow

6022 yellow, 2001 green

3.01 : 1

Purple × white (flowers)

Purple

705 purple, 224 white

3.15 : 1

Inflated × constricted (pods)

Inflated

882 inflated, 299 constricted

2.95 : 1

Green × yellow (unripe pods)

Green

428 green, 152 yellow

2.82 : 1

Axial × terminal (flower position)

Axial

651 axial, 207 terminal

3.14 : 1

Long × short (stems)

Long

787 long, 277 short

2.84 : 1

The 3 : 1 ratio observed in the F2 generation is the key to understanding the mechanism of genetic transmission. In the next section, we use PCR analysis of the W and w alleles to explain why this ratio is produced.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The Principle of Segregation The 3 : 1 ratio can be explained with reference to FIGURE 3.4. This is the heart of Mendelian genetics. You should master it and be able to use it to deduce the progeny types produced in crosses. The diagram illustrates these key features of single-gene inheritance: 1. Genes come in pairs, which means that a cell or individual has two copies (alleles) of each gene. 2. For each pair of genes, the alleles may be identical (homozygous WW or homozygous ww), or they may be different (heterozygous Ww).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3. Each reproductive cell (gamete) produced by an individual contains only one allele of each gene (that is, either W or w). 4. In the formation of gametes, any particular gamete is equally likely to include either allele (hence, from a heterozygous Ww genotype, half the gametes contain W and the other half

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

contain w). 5. The union of male and female reproductive cells is a random process that reunites the alleles in pairs.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.4 A diagrammatic explanation of the 3 : 1 ratio of dominant : recessive morphological phenotypes observed in the F2 generation of a monohybrid cross. (A) Two true-breeding parental strains, one round and one wrinkled, are crossed and yield roundseeded F1 progeny. (B) Those progeny are self-fertilized to yield the F2 generation. A 3 : 1 ratio of phenotypes is observed in the F2 generation because of dominance. Note that the ratio of WW : Ww : ww genotypes in the F2 generation is 1 : 2 : 1, as can be seen from the restriction fragment phenotypes.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

The essential feature of transmission genetics is the separation (technically called segregation), in unaltered form, of the two alleles in an individual during the formation of its reproductive cells. Segregation corresponds to points 3 and 4 in the foregoing list. The principle of segregation is sometimes called Mendel's first law. Principle of segregation: In the formation of gametes, the paired hereditary determinants separate (segregate) in such a way that each gamete is equally likely to contain either member of the pair. Another key feature of transmission genetics is that the hereditary determinants are present as pairs in both the parental organisms and the progeny organisms but as single copies in the reproductive cells. This feature corresponds to points 1 and 5 in the foregoing list.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Figure 3.4 illustrates the biological mechanism underlying the important Mendelian ratios in the F2 generation of 3 : 1 for phenotypes and 1 : 2 : 1 for genotypes. To understand these ratios, consider first the parental generation in which the original cross is WW × ww (Figure 3.4A). The sex of the parents is not stated because reciprocal crosses yield identical results. (There is, however, a convention in genetics that unless otherwise specified, crosses are given with the female parent listed first.) In the original cross, the WW parent produces only W-containing gametes, whereas the ww parent produces only w-containing gametes. Segregation still takes place in the homozygous genotypes as well as in the heterozygous genotype, even though all of the gametes carry the same type of allele (W from homozygous WW and w from homozygous ww). When the W-bearing and w-bearing gametes come together in fertilization, the hybrid genotype is

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

heterozygous Ww, which is shown by the bands in the gel icon next to the F1 progeny. With regard to seed shape, the hybrid Ww seeds are round because W is dominant over w. Now consider Figure 3.4B. When the heterozygous F1 progeny form gametes, segregation implies that half the gametes will contain the W allele and the other half will contain the w allele. These gametes come together at random when an F1 individual is self-fertilized or when two F1 individuals are crossed. The result of

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

random fertilization can be deduced from the sort of crossmultiplication square shown at the bottom of the figure, in which the female gametes and their frequencies are arrayed across the top margin and the male gametes and their frequencies along the lefthand margin. This calculating device, which is widely used in genetics, is called a Punnett square after its inventor Reginald C. Punnett (1875–1967). The Punnett square in Figure 3.4B shows that random combinations of the F1 gametes result in an F2 generation with the genotypic composition 1/4 WW, 1/2 Ww, and 1/4 ww. This can be confirmed by the PCR banding patterns in the gel icons because of the codominance of W and w with respect to the molecular phenotype. But because W is dominant over w with respect to the morphological phenotype, the WW and Ww genotypes have round seeds and the ww genotypes have wrinkled seeds, yielding the phenotypic ratio of round : wrinkled seeds of 3 : 1. Hence, it is a combination of segregation, random union of gametes, and dominance that results in the 3 : 1 ratio.

ROOTS OF DISCOVERY What Did Gregor Mendel Think He Discovered? Gregor Mendel (1866) Monastery of St. Thomas, Brno [then Brünn], Czech Republic

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Experiments on Plant Hybrids (original in German) Mendel's paper is remarkable and is worth reading. Although he is given credit for discovering segregation, he never uses this term. His description of segregation is found in the first passage in italics in the excerpt. (All of the text not in italics are reproduced from the original.) In his description of the process, Mendel takes us carefully through the separation of A and a in gametes and their coming together again at random in fertilization. One flaw in the description is Mendel's occasional confusion between genotype and phenotype, which is illustrated by his writing A instead of AA and a instead of aa in the display toward the end of the passage. Most early geneticists made no consistent distinction between genotype and phenotype until 1909, when the terms themselves were coined.

“Whether the plan by which the individual experiments were set up and carried out was adequate to the assigned task should

”

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

be decided by a benevolent judgment.

Artificial fertilization undertaken on ornamental plants to obtain new color variants initiated the experiments reported here. The striking regularity with which the same hybrid forms always reappeared whenever fertilization between like species took place suggested further experiments whose task it was to follow that development of hybrids in their progeny…. This paper discusses the attempt at such a detailed experiment…. Whether the plan by which the individual experiments were set up and carried out was adequate to the assigned task should be decided by a benevolent judgment….

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

[Here the experimental results are described in detail.] Thus, experimentation also justifies the assumption that pea hybrids form germinal and pollen cells that in their composition correspond in equal numbers to all the constant forms resulting from the combination of traits united through fertilization. The difference of forms among the progeny of hybrids, as well as the ratios in which they are observed, find an adequate explanation in the principle [of segregation] just deduced. The simplest case is given by the series for one pair of differing traits. It is shown

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

that this series is described by the expression: A + 2Aa + a, in which A and a signify the forms with constant differing traits, and Aa the form hybrid for both. The series contains four individuals in three different terms. In their production, pollen and germinal cells of form A and a participate, on the average, equally in fertilization; therefore each form manifests itself twice, since four individuals are produced. Participating in fertilization are thus:

It is entirely a matter of chance which of the two kinds of pollen combines with each single germinal cell. However, according to the laws of probability, in an average of many cases it will always happen that every pollen form A and a will unite equally often with every germinal-cell form A and a; therefore, in fertilization, one of the two pollen cells A will meet a germinal cell A, the other a germinal cell a, and equally, one pollen cell a will become associated with a germinal cell A, and the other a.

The result of fertilization can be visualized by writing the designations for associated germinal and pollen cells in the form

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

of fractions, pollen cells above the line, germinal cells below. In the case under discussion one obtains

In the first and fourth terms germinal and pollen cells are alike;

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

therefore the products of their association must be constant, namely A and a. In the second and third, however, a union of the two differing parental traits takes place again; therefore the forms arising from such fertilizations are absolutely identical with the hybrid from which they derive. Thus, repeated hybridization takes place. The striking phenomenon, that hybrids are able to produce, in addition to the two parental types, progeny that resemble themselves, is thus explained: Aa and aA both give the same association, Aa, since, as mentioned earlier, it makes no difference to the consequence of fertilization which of the two traits belongs to the pollen and which to the germinal cell. Therefore,

This represents the average course of self-fertilization of hybrids when two differing traits are associated in them. In individual flowers and individual plants, however, the ratio in which the members of the series are formed may be subject to not insignificant deviations…. Thus, it was proven experimentally that, in Pisum, hybrids form different kinds of germinal and pollen cells and that this is the reason for the variability of their offspring. Source: G. Mendel, Verhandlungen des naturforschenden den Vereines in Brünn 4 (1866): 3–47.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

The ratio of F2 genotypes is as important as the ratio of F2 phenotypes. The Punnett square in Figure 3.4B also shows that the ratio of WW : Ww : ww genotypes is 1 : 2 : 1, which can be confirmed directly by the PCR analysis.

Verification of Segregation The round seeds in Figure 3.4B conceal a genotypic ratio of 1 WW : 2 Ww. Put another way, among the F2 seeds that are round (or, more generally, among organisms that show the dominant morphological phenotype), 1/3 are homozygous (in this example, WW) and 2/3 are heterozygous (in this example, Ww). This conclusion is obvious from the PCR patterns in Figure 3.4B, but it is not at all obvious from the morphological phenotypes. Unless you knew something about genetics already, it would be a very bold hypothesis, because it implies that two organisms with the same morphological phenotype (in this case, round seeds) might nevertheless differ in molecular phenotype and in genotype.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Yet this is exactly what Mendel proposed. But how could this hypothesis be tested experimentally? Mendel realized that it could be tested via self-fertilization of the F2 plants. With self-fertilization, plants grown from the homozygous WW genotypes should be truebreeding for round seeds, whereas those from the heterozygous Ww genotypes should yield round and wrinkled seeds in the ratio of 3 : 1. In contrast, the plants grown from wrinkled seeds should be true-breeding for wrinkled because these plants are homozygous ww. The results Mendel obtained are summarized in FIGURE 3.5. As predicted from the genetic hypothesis, plants grown from F2 wrinkled seeds were true-breeding for wrinkled seeds, yielding only wrinkled seeds in the F3 generation. But some of the plants grown

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

from round seeds showed evidence of segregation. Among 565 plants grown from round F2 seeds, 372 plants produced both round and wrinkled seeds in a proportion very close to 3 : 1, whereas the remaining 193 plants produced only round seeds in the F3 generation. The ratio 193 : 372 equals 1 : 1.93, which is very close to the ratio 1 : 2 of WW : Ww genotypes predicted from the genetic hypothesis.

FIGURE 3.5 Summary of F2 phenotypes and the progeny produced by self-fertilization.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

An important feature of the homozygous round and homozygous wrinkled seeds produced in the F2 and F3 generations is that the phenotypes are exactly the same as those observed in the original parents in the P1 generation. This makes sense in terms of DNA, because the DNA of each allele remains unaltered unless a new mutation happens to occur. Mendel described this result in a letter by saying that in the progeny of crosses, “the two parental traits appear, separated and unchanged, and there is nothing to indicate that one of them has either inherited or taken over anything from the other.” From this finding, he concluded that the hereditary determinants for the traits in the parental lines were transmitted as

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

two different elements that retain their purity in the hybrids. In other words, the hereditary determinants do not “mix” or “contaminate” each other. In modern terminology, this means that, with rare but important exceptions, genes are transmitted unchanged from generation to generation. In fact, this concept, sometimes referred to as the particulate gene, is considered by many to be Mendel's most important contribution.

The Testcross and the Backcross

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Another straightforward way of testing the genetic hypothesis in Figure 3.4 is by means of a testcross, a cross between an organism that is heterozygous for one or more genes (for example, Ww) and an organism that is homozygous for the recessive allele (for example, ww). The result of such a testcross is shown in FIGURE 3.6. Because the heterozygous parent is expected to produce W and w gametes in equal numbers, whereas the homozygous recessive produces only w gametes, the expected progeny are 1/2 with the genotype Ww and 1/2 with the genotype ww. The former have the dominant phenotype because W is dominant over w, and the latter have the recessive phenotype. A testcross is often extremely useful in genetic analysis: In a testcross, the phenotypes of the progeny reveal the relative frequencies of the different gametes produced by the heterozygous parent, because the recessive parent contributes only recessive alleles. Furthermore, in contrast to the case of F2 progeny from a dihybrid cross, knowing the phenotypes allows one to precisely infer the genotype.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

FIGURE 3.6 In a testcross of a Ww heterozygous parent with a ww homozygous recessive, the progeny are Ww and ww in the ratio of 1 : 1. A testcross shows the result of segregation.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Mendel carried out a series of testcrosses with various traits. The results are shown in TABLE 3.2. In all cases, the ratio of phenotypes among the testcross progeny is very close to the 1 : 1 ratio expected from segregation of the alleles in the heterozygous parent.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

TABLE 3.2 Mendel's testcross results Testcross (F1 heterozygote × homozygous

Progeny from

recessive)

testcross

Round × wrinkled seeds

193 round, 192

1.01 :

wrinkled

1

196 yellow, 189 green

1.04 :

Yellow × green seeds

Ratio

1 Purple × white flowers

85 purple, 81 white

1.05 : 1

Long × short stems

87 long, 79 short

1.10 : 1

Another valuable type of cross is a backcross, in which hybrid organisms are crossed with one of the parental genotypes. Backcrosses are commonly used by geneticists and by plant and animal breeders. Note that the testcrosses in Table 3.2 are also backcrosses, because in each case, the F1 heterozygous parent

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

came from a cross between the homozygous dominant and the homozygous recessive.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

SUMMING UP ■ Mendel's experiments began with crossing of true-breeding strains (P1 generation) that differed for a particular phenotype to create hybrids (F1). Hybrids were then crossed and the resulting progeny (F2) were analyzed. ■ The 3 : 1 ratio observed in the F2 generation led Mendel to infer the principle of segregation. ■ Mendel confirmed the principle of segregation by showing that in F2 progeny, 2/3 of plants with the dominant phenotype are homozygous and the remaining 1/3 are heterozygous.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The frequencies of gametes produced by a heterozygote can be directly determined with a testcross.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.3 Segregation of Two or More Genes The results of many genetic crosses depend on the segregation of the alleles of two or more genes. The genes may be in different chromosomes or in the same chromosome. Although in this section we consider the case of genes that are in two different chromosomes, the same principles apply to genes that are in the same chromosome but are so far distant from each other that they

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

segregate independently. The case of linkage of genes in the same chromosome is examined in the Genetic Linkage and Chromosome Mapping chapter. To illustrate the principles, we consider again a cross between homozygous genotypes, but in this case homozygous for the alleles of two genes. A specific example is a true-breeding variety of garden peas with seeds that are wrinkled and green (genotype ww gg) versus a variety with seeds that are round and yellow (genotype WW GG). As suggested by the use of uppercase and lowercase symbols for the alleles, the dominant alleles are W and G, the recessive alleles w and g. The mutant gene responsible for Mendel's green seeds has been identified. It is an inborn error in metabolism that blocks the metabolic pathway from breaking down the green pigment chlorophyll. Homozygous mutant seeds are unable to break down their chlorophyll and therefore stay green, whereas wild-type seeds do break down their chlorophyll and turn yellow, like the leaves of certain trees in the autumn. The mutant gene is officially named staygreen. A cross of WW GG plants with ww gg plants yields F1 seeds with the genotype Ww Gg, which are phenotypically round and yellow because of the dominance relations. When the F1 seeds are grown into mature plants and self-fertilized, the F2 progeny show the result of simultaneous segregation of the W, w allele pair and the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

G, g allele pair. When Mendel performed this cross, he obtained the following numbers of F2 seeds:

In these data, the first thing to be noted is the expected 3 : 1 ratio for each trait considered separately. The ratio of round : wrinkled (pooling across yellow and green) is:

The ratio of yellow : green (pooling across round and wrinkled) is:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Both of these ratios are in satisfactory agreement with 3 : 1. (Testing for goodness of fit to a predicted ratio is described in the chapter titled The Chromosomal Basis of Inheritance). Furthermore, in the F2 progeny of the dihybrid cross, the separate 3 : 1 ratios for the two traits were combined at random. With random combinations, as shown in FIGURE 3.7, among the 3/4 of the progeny that are round, 3/4 will be yellow and 1/4 green; similarly, among the 1/4 of the progeny that are wrinkled, 3/4 will be yellow and 1/4 green. The overall proportions of round yellow to round green to wrinkled yellow to wrinkled green are, therefore, expected to be:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

The observed ratio of 315 : 108 : 101 : 32 equals 9.84 : 3.38 : 3.16 : 1, which is a satisfactory fit to the 9 : 3 : 3 : 1 ratio expected from the Punnett square in Figure 3.7.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.7 The 3 : 1 ratio of round : wrinkled, when combined at random with the 3 : 1 ratio of yellow : green, yields the 9 : 3 : 3 : 1 ratio observed in the F2 progeny of the dihybrid cross.

The Principle of Independent Assortment The independent segregation of the W, w and G, g allele pairs is illustrated in FIGURE 3.8. What independence means is that if a gamete contains W, it is equally likely to contain G or g; and if a gamete contains w, it is equally likely to contain G or g. The implication is that the four gametes are formed in equal frequencies:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

FIGURE 3.8 Independent segregation of the W, w and G, g allele pairs means that among each of the W and w gametic classes, the ratio of G : g is 1 : 1. Likewise, among each of the G and g gametic classes, the ratio of W : w is 1 : 1.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The result of independent assortment when the four types of gametes combine at random to form the zygotes of the next generation is shown in FIGURE 3.9. Note that the expected ratio of phenotypes among the F2 progeny is:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.9 Independent assortment is the biological basis for the 9 : 3 : 3 : 1 ratio of F2 phenotypes resulting from a dihybrid cross. (A) A cross of true-breeding round yellow and wrinkled green parental strains yields round, yellow F1 progeny. (B) The results of self-fertilizing the F1 progeny are shown as a Punnett square.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

However, as the Punnett square also shows, the ratio of genotypes in the F2 generation is more complex. The reason for this ratio is shown in FIGURE 3.10. Among seeds that have the WW genotype, the ratio of

FIGURE 3.10 Genotypes and phenotypes of the F2 progeny of the dihybrid cross for seed shape and seed color.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Among seeds that have the Ww genotype, the ratio is:

(which is a 1 : 2 : 1 ratio multiplied by 2, because there are twice as many Ww genotypes as either WW or ww). Among seeds that have the ww genotype, the ratio of

Combining these gives an overall genotypic ratio of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

The phenotypes of the seeds are shown beneath the genotypes, and the combined phenotypic ratio is:

The principle of independent segregation of two pairs of alleles in different chromosomes (or located sufficiently far apart in the same chromosome) has come to be known as the principle of independent assortment. It is also sometimes referred to as Mendel's second law. Principle of independent assortment: Segregation of the members of any pair of alleles is independent of the segregation of other pairs in the formation of reproductive cells.

The Testcross with Independently Assorting Genes Thus, in F2 progeny, when one of each segregating allele is

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

dominant, nine different genotypes appear as only four distinguishable phenotypes. This makes the results of such crosses difficult to interpret. As an alternative, the testcross can be used:

The result is shown in FIGURE 3.11. Because plants with doubly heterozygous genotypes produce four types of gametes—W G, W g, w G, and w g—in equal frequencies, whereas the plants with ww gg genotypes produce only w g gametes, the possible progeny genotypes are Ww Gg, Ww gg, ww Gg, and ww gg, and these are expected in equal frequencies. Because of the dominance relations —W over w and G over g—the progeny phenotypes are expected

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

to be round yellow, round green, wrinkled yellow, and wrinkled green in a ratio of 1 : 1 : 1 : 1.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.11 Genotypes and phenotypes resulting from a testcross of the Ww Gg double heterozygote.

As with the one-gene testcross, in a two-gene testcross the ratio of progeny phenotypes is a direct demonstration of the ratio of gametes produced by the doubly heterozygous parent, meaning that there is a 1 : 1 correspondence between genotype and phenotype. In his actual cross, Mendel obtained 55 round yellow, 51 round green, 49 wrinkled yellow, and 53 wrinkled green progeny, which is in good agreement with the predicted 1 : 1 : 1 : 1 ratio.

Three or More Genes A Punnett square of the type in Figure 3.9 is a nice way to show the logic behind the genotype and phenotype frequencies for two genes that undergo independent assortment. As a method for

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

solving problems, however, it is not very efficient. In working a problem, especially when time is limited (during an exam, for example), drawing and filling out the whole square takes too long and provides many opportunities to make mistakes. As an alternative, we can use the forked line approach. To illustrate this method, consider an example of a trihybrid cross with the allele pairs (W, w), (G, g), and (P, p), where P is a dominant allele for purple flowers and p is a recessive mutation for

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

white flowers. A Punnett square is a cube containing 64 cells that is tricky to draw and tedious to complete. If the three genes assort independently, however, then instead of trying to analyze the assortment of all three loci simultaneously, we can consider how the genes segregate individually, and then combine those results. Furthermore, we will consider only what we can see—the phenotypes. This approach is illustrated in FIGURE 3.12, which is called a forked-line diagram. The underlying principle is simple: If we were looking at the three genes one at a time, we would expect to observe 3 : 1 ratios of the dominant phenotype to the recessive phenotype. Thus, in the diagram, we first consider the phenotypes associated with the W locus, where we would expect to see 3/4 of the progeny to be round and 1/4 wrinkled. Then, having separated the progeny by the first phenotype, we would ask what the expected ratio of yellow to green seeds should be for the roundversus wrinkled-seeded plants. For each, and assuming independent assortment, that ratio will be 3/4 : 1/4. Finally, for each of the resulting four phenotypic classes, we separate them into the expected ratio—3/4 : 1/4—for purple versus white flowers, resulting in eight phenotypic classes.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

FIGURE 3.12 A forked line diagram demonstrating how phenotypic ratios can be predicted in a trihybrid cross, without resorting to use of a Punnett square.

So what are the expected ratios of those eight classes? They can be obtained by multiplying the fractions along each of the particular paths leading to one of the class. For example, the fraction of the progeny expected to be round, yellow, and purple is:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

What we are doing is to ask the question, “What fraction of the progeny would we expect to be round (3/4) and wrinkled (3/4) and yellow (3/4)?” We can then do the same for the remaining seven phenotypic categories. The results of this analysis are shown in TABLE 3.3. It is worth taking some time to consider this table carefully. In cases of phenotypes resulting from dominant genotypes, the genotype is given as the dominant allele followed by a dash (for example, W−). This indicates that we do not know what the other allele of the genotype is—it could be either the dominant allele or the recessive allele.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

TABLE 3.3 Independent assortment in a trihybrid cross Phenotype

Genotype

Expected

Observed

Expected

27/64

269

270

fraction Round, yellow, purple

W— G— P —

Round, yellow, white

W— G— pp

3/64

98

90

Round, green, purple

W— gg P—

9/64

86

90

Round, green, white

W— gg pp

3/64

27

30

Wrinkled, yellow,

ww G— P—

9/64

88

90

ww G— pp

3/64

34

30

ww gg P—

3/64

30

30

ww gg pp

1/64

7

10

purple Wrinkled, yellow, white Wrinkled, green, purple Wrinkled, green, white Phenotypes expected in a trihybrid cross. A dash used in a genotype symbol indicates that either the dominant or the recessive allele is present; for example, W—refers

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

collectively to the genotypes WW and Ww. (The expected numbers total 640 rather than 639 because of round-off error.)

Mendel actually carried out this three-gene cross. The results for the phenotypes, shown in Table 3.3, conform well to the ratio of 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1 expected from independent assortment. The eight progeny types represent 33 = 27 different genotypes. Mendel did a testcross for each offspring having the dominant phenotype for one or more traits so as to determine whether its genotype was homozygous or heterozygous for each of the genes. This effort

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

alone required 632 testcrosses. Little wonder that he complained that “of all the experiments, [this one] required the most time and effort.” We have emphasized the efficiency and utility of a more conceptual approach to genetic calculations than the Punnett square. We will come back to this issue when we consider how probability theory can be more formally applied to genetic reasoning. Before doing so, however, let us address the question of genetic analysis of human heredity.

SUMMING UP ■ In dihybrid crosses involving genes that affect two different traits, each gene follows the principle of segregation. ■ The 9 : 3 : 3 : 1 ratio of phenotypes observed in the F2 generation can be explained by the principle of independent assortment. ■ Testcrosses produce a 1 : 1 correspondence of genotype and phenotype in the progeny.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The principle of independent assortment can be applied to crosses involving three or more genes, most efficiently with the forked line approach.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.4 Human Pedigree Analysis Large deviations from expected genetic ratios are often found in individual human families and in domesticated large animals because of the relatively small number of progeny. The effects of segregation are nevertheless evident upon examination of the phenotypes among several generations of related individuals. A diagram of a family tree showing the phenotype of each individual among a group of relatives is a pedigree. In this section, we introduce basic concepts in pedigree analysis.

Characteristics of Dominant and Recessive

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Inheritance FIGURE 3.13 defines the standard symbols used in depicting human pedigrees. Females are represented by circles and males by squares. (A diamond is used if the sex is unknown—as, for example, in a miscarriage.) Persons with the phenotype of interest are indicated by colored or shaded symbols. For recessive alleles, heterozygous carriers are sometimes depicted with half-filled symbols. A mating between a female and a male is indicated by joining their symbols with a horizontal line, which is connected vertically to a second horizontal line, below, that connects the symbols for their offspring. The offspring within a sibship, called siblings or sibs, are represented from left to right in order of their birth.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.13 Conventional symbols used in depicting human pedigrees.

A pedigree for the trait Huntington disease, caused by a dominant mutation, is shown in FIGURE 3.14. The numbers in the pedigree are provided for convenience in referring to particular persons. The successive generations are designated by Roman numerals. Within any generation, all of the persons are numbered consecutively from left to right. The pedigree starts with the woman I-1 and the man I2. The man has Huntington disease, which is a progressive nerve degeneration that usually begins about middle age. It results in severe physical and intellectual disability, and ultimately in death. The dominant allele, HD, that causes Huntington disease is rare. All affected persons in the pedigree have the heterozygous genotype HD hd, whereas nonaffected persons have the homozygous normal genotype hd hd. In many families, the disease has complete penetrance. The penetrance of a genetic disorder is the proportion of individuals with the at-risk genotype who express the trait; complete penetrance means the trait is expressed in 100 percent of persons with that genotype. The pedigree demonstrates the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

following characteristic features of inheritance due to a rare dominant allele with complete penetrance: 1. Females and males are equally likely to be affected. 2. Affected offspring have one affected parent (except for rare new mutations), and the affected parent is equally likely to be the mother or the father. 3. On average, half of the individuals in sibships with an affected parent are affected.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.14 Pedigree of a human family showing the inheritance of the dominant gene for Huntington disease. Females and males are represented by circles and squares, respectively. Red symbols indicate persons affected with the disease.

A pedigree for a trait due to a homozygous recessive allele is shown in FIGURE 3.15. The trait is albinism, absence of pigment in the skin, hair, and iris of the eyes. This pedigree illustrates characteristics of inheritance due to a rare recessive allele with complete penetrance: 1. Females and males are equally likely to be affected. 2. Affected individuals, if they reproduce, usually have unaffected progeny.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3. Most affected individuals have unaffected parents. 4. The parents of affected individuals are often relatives. 5. Among siblings of affected individuals, the proportion affected is approximately 25 percent.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.15 Pedigree of albinism. With recessive inheritance, affected persons (filled symbols) often have unaffected parents. The double horizontal line indicates a mating between relatives—in this case, first cousins.

With rare recessive inheritance, the mates of homozygous affected persons are usually homozygous for the normal allele, so all of the offspring will be heterozygous and not affected. Heterozygous carriers of the mutant allele are considerably more common than homozygous affected individuals, because it is more likely that a person will inherit only one copy of a rare mutant allele than two copies. Most homozygous recessive genotypes therefore result from matings between carriers (heterozygous × heterozygous), in which each offspring has a 1/4 chance of being affected. Another important feature of rare recessive inheritance is that the parents of affected individuals are often related (consanguineous). In a pedigree, a mating between relatives is indicated with a double line connecting the partners, as for the first-

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

cousin mating in Figure 3.15. Mating between relatives is important for recessive alleles to become homozygous; that is, when a recessive allele is rare, it is more likely to become homozygous through inheritance from a common ancestor than from parents who are completely unrelated. The reason for this pattern is that the carrier of a rare allele may have many descendants who are also carriers. If two of these carriers should mate with each other (for example, in a first-cousin mating), then the hidden recessive allele can become homozygous with a probability of 1/4. Mating between relatives constitutes inbreeding, and the consequences of inbreeding are discussed further in the Genes in Populations chapter. Because an affected individual indicates that the parents are heterozygous carriers, the expected proportion of affected individuals among the siblings is approximately 25 percent, but the exact value depends on the details of how affected individuals are identified and included in the analysis.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Most Human Genetic Variation Is Not “Bad” Before the advent of molecular methods, there were many practical obstacles to the study of human genetics. With the exception of traits such as the ABO and other blood groups, few traits showing simple Mendelian inheritance were known. Most of these were associated with genetic diseases, and these presented special challenges: ■ Most genes that cause simple Mendelian genetic diseases are rare, so they are observed in only a small number of families. ■ Many genes for simple Mendelian diseases are recessive, so they are not detected in heterozygous genotypes. ■ The number of offspring per human family is relatively small, so segregation cannot usually be detected in single sibships.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

■ The human geneticist cannot perform testcrosses, backcrosses, or other experimental matings. This text includes numerous examples of how molecular genetics has revolutionized the study of human genetics. For example, in the chapter DNA Structure and Genetic Variation, we discussed the prevalence of single-nucleotide polymorphisms (SNPs) and copynumber variation (CNV) in the human genome. Only a very small number of these common polymorphisms are associated with diseases, and even those serve as genetic risk factors that interact with other risk factors, including other genes and the environment, to determine whether a person actually develops the disease. Although most SNPs and CNVs have no adverse effects, both types of variation show simple Mendelian inheritance, which is why human geneticists have come to rely on them for family and population studies.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Nevertheless, a few, seemingly harmless, simple-Mendelian traits were detected in the pre-molecular era. One of the best known was associated with the ability to taste a chemical substance known as phenylthiocarbamide (PTC), which has the molecular structure shown here:

PTC is an artificial chemical created by an industrial chemist in the early 1930s. The taste polymorphism was discovered one day when he carelessly released a cloud of powdered PTC into the air. The PTC powder did not bother the chemist at all, but his lab mate loudly complained about the bitter taste it left in his mouth. Out of curiosity, the chemist started to test family and friends for their ability to taste PTC, and he recruited a geneticist who began to

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

study the situation. It was soon shown from family studies that the ability to taste PTC is a trait inherited as a simple Mendelian dominant. In European populations, approximately 70 percent of people are tasters and 30 percent are nontasters, but these proportions differ greatly among ethnic groups. Among people of African or Asian origin, the frequency of tasters is approximately 90 percent, whereas among Australian aborigines it is only 50 percent. The ability to taste PTC varies, however. The most sensitive

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

tasters can taste concentrations as low as 0.001 millimolar (mM), whereas the most insensitive nontasters fail to detect concentrations as high as 10 mM. For classifying individuals as “tasters” or “nontasters,” an arbitrary cutoff is employed, typically at a concentration of 0.2 mM PTC. Most of the variation in tasting ability between tasters and nontasters is due to the major taster polymorphism, but some differences are due to other genes, gender, and probably environmental factors. The result of the other variables is that about 5 percent of the heterozygous tasters are classified as “nontaster” and at least 5 percent of the homozygous nontasters become classified as “tasters.” The molecular basis of the taster polymorphism is now known to reside in a taste receptor protein known as hTAS2R38. Several alleles of the gene exist, but the most common forms of the protein differ by three amino acids at scattered positions along the protein. The allelic forms are known as PAV and AVI, because the three key amino acids in the PAV protein are proline, alanine, and valine, whereas these three positions in the AVI protein are occupied by alanine, valine, and isoleucine. The PAV form is the one that confers the ability to taste PTC. When you think about it, a polymorphism in PTC tasting makes no sense. PTC is a completely artificial chemical synthesized in the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

laboratory, so why should there be a polymorphism in the ability to taste it? A clue comes from the observation that the chemical structure of PTC resembles a large and heterogeneous class of molecules called glucosinolates. These compounds are synthesized by some plants, including some human food plants, and their synthesis likely evolved as a chemical defense against plant-eating insects. Among the plants that produce glucosinolates is one singled out by former President George H. W. Bush, who in 1989 removed broccoli from the White House menu, proclaiming: “I do not like broccoli. And I haven't liked it since I was a little kid and my mother made me eat it. And I'm president of the United States and I'm not going to eat any more broccoli!” In good-humored protest, broccoli growers throughout the United States sent Bush tons of the stuff. Seventeen years later, new studies showed that individuals carrying the PAV form of the hTAS2R38 taste receptor do, in fact, find broccoli to be significantly more bitter tasting than individuals homozygous for the allele encoding the AVI form. Tasters also report a greater perceived bitterness of collard greens, turnips, rutabagas, and horseradish.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Molecular Markers in Human Pedigrees Because techniques for manipulating DNA allow direct access to the DNA, modern genetic studies of human pedigrees are carried out primarily using genetic markers present in the DNA itself, rather than through the phenotypes produced by mutant genes. Various types of DNA polymorphisms were discussed in the DNA Structure and Genetic Variation chapter, along with the methods by which they are detected and studied. An example of a DNA polymorphism segregating in a three-generation human pedigree is shown in FIGURE 3.16. The type of polymorphism is a simple sequence repeat polymorphism (SSRP), in which each allele differs in size according to the number of copies it contains of a short DNA

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

sequence repeated in tandem. The differences in size are detected by electrophoresis after amplification of the region by PCR. SSRPs usually have many codominant alleles, and the majority of individuals are heterozygous for two different alleles. In the example in Figure 3.16, each of the parents is heterozygous, as are all of the children.

FIGURE 3.16 Human pedigree showing segregation of SSRP alleles. Six alleles (A1–A6) are present in the pedigree, but any one person can have only one allele (if homozygous) or two alleles (if heterozygous).

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Six alleles are depicted in Figure 3.16, denoted by A1 through A6. In the gel, the numbers of the bands correspond to the subscripts of the alleles. The mating in generation II is between two heterozygous genotypes: A4A6 × A1A3. Because of segregation in each parent, four genotypes are possible among the offspring (A4A1, A4A3, A6A1, and A6A3); these would conventionally be written with the smaller subscript first, as A1A4, A3A4, A1A6, and A3A6. With random fertilization, the offspring genotypes are equally likely, as may be verified from a Punnett square for the mating. Figure 3.16 illustrates some of the principal advantages of multiple, codominant alleles for human pedigree analysis:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

■ Heterozygous genotypes can be distinguished from homozygous genotypes. ■ Many individuals in the population are heterozygous, and so many matings are informative in regard to segregation. ■ Each segregating genetic marker yields up to four distinguishable offspring genotypes.

SUMMING UP ■ Inheritance in humans can be analyzed by means of pedigree analysis. ■ In the case of dominant traits, affected individuals typically have affected parents. ■ In the case of recessive traits, affected individuals typically have unaffected parents. ■ Heterozygotes for recessive traits are referred to as carriers and are more common in populations than homozygotes with the trait.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Simple Mendelian genetic diseases are rare.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.5 Incomplete Dominance and Epistasis Dominance and codominance are not the only possibilities for pairs of alleles. There are situations of incomplete dominance, in which the phenotype of the heterozygous genotype is intermediate between the phenotypes of the homozygous genotypes. A classic example of incomplete dominance concerns flower color in the snapdragon Antirrhinum majus (FIGURE 3.17). In wild-type flowers, a red type of anthocyanin pigment is formed by a sequence of enzymatic reactions. A wild-type enzyme, encoded by the I allele, is the rate-limiting factor for the overall reaction, so the amount of red pigment is determined by the amount of enzyme that the I allele produces. The alternative i allele codes for an inactive enzyme, and ii flowers are ivory in color. Because the amount of the critical enzyme is reduced in Ii heterozygotes, the amount of red pigment in the flowers is reduced as well, and the effect of the dilution is to make the flowers pink. A cross between plants differing in flower color therefore gives direct phenotypic evidence of segregation Figure 3.21). The cross II (red) × ii (ivory) yields F1 plants with genotype Ii and pink flowers. In the F2 progeny obtained by self-pollination of the F1 hybrids, one experiment resulted in 22

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

plants with red flowers, 52 with pink flowers, and 23 with ivory flowers, which fits the expected ratio of 1 : 2 : 1.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.17 Red versus white flower color in snapdragons shows incomplete dominance.

ROOTS OF DISCOVERY This Land Is Your Land The Huntington's Disease Collaborative Research Group (1993) Comprising 58 authors among 9 institutions

A Novel Gene Containing a Trinucleotide Repeat That Is

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Expanded and Unstable on Huntington's Disease Chromosomes Modern genetic research is sometimes carried out by large collaborative groups in a number of research institutions scattered across several countries. This approach is exemplified by the search for the gene responsible for Huntington disease. The search was highly publicized because of the severity of the disease, the late age of onset, and the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

dominant inheritance. Famed folk singer Woody Guthrie, who wrote “This Land Is Your Land” and other well-known tunes, died of the disease in 1967. When the gene was identified, it turned out to encode a protein (now called huntingtin) of unknown function that is expressed in many cell types throughout the body and not, as expected, exclusively in nervous tissue. Within the coding sequence of this gene is a trinucleotide repeat (5′-CAG-3′) that is repeated in tandem a number of times according to the general formula (5′-CAG-3′)n. Among normal alleles, the number n of repeats ranges from 11 to 34, with an average of 18; among mutant alleles, the number of repeats ranges from 40 to 86. This tandem repeat is genetically unstable in that it can, by some unknown mechanism, increase in copy number (“expand”). In two cases in which a new mutant allele was analyzed, one had increased in repeat number from 36 to 44 and the other from 33 to 49. This mutational mechanism is quite common in some human genetic diseases. The excerpt cites several other examples. The authors also emphasize that their discovery raises important ethical issues, including those related to genetic testing, confidentiality, and informed consent.

“We consider it of the utmost importance that the current Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

“We consider it of the utmost importance that the current internationally accepted guidelines and counseling protocols for testing people at risk continue to be observed, and that samples from unaffected relatives should not be tested inadvertently or

”

without full consent.

The genetic defect causing HD was assigned to chromosome 4

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

in one of the first successful linkage analyses using DNA markers in humans. Since that time, we have pursued an approach to isolating and characterizing the HD gene based on progressively refining its localization…. [We have found that a] 500-kb segment is the most likely site of the genetic defect. [The abbreviation “kb” stands for kilobase pairs; 1 kb equals 1000 base pairs.] Within this region, we have identified a large gene, spanning approximately 210 kb, that encodes a previously undescribed protein. The reading frame contains a polymorphic (CAG)n trinucleotide repeat with at least 17 alleles in the normal population, varying from 11 to 34 CAG copies. On HD chromosomes, the length of the trinucleotide repeat is substantially increased…. It can be expected that the capacity to monitor directly the size of the trinucleotide repeat in individuals “at risk” for HD will revolutionize testing for the disorder…. The authors then go on to point out the importance of adhering to rigorous ethical standards in testing programs—in particular with respect to testing unaffected relatives of affected individuals without their fully informed consent. In the case of Huntington disease, this is particularly important, since the disease-associated alleles are dominant and do not manifest themselves phenotypically until later in life. Thus, for example, a seemingly healthy young adult could learn as a result of testing that she or he would develop this debilitating and

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

ultimately fatal disease later in life—information that some would rather not know. Source: MacDonald, ME, Ambrose, CM, Duyao, MP. A novel gene containing a trinucleotide repeat that is expanded and unstable on Huntington's disease chromosomes. The Huntington's Disease Collaborative Research Group, Cell 72 (1993): 971–983.

Multiple Alleles The occurrence of multiple alleles is exemplified by the alleles A1–A6 of the SSR marker in the human pedigree in Figure 3.16.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Multiple alleles are relatively common in natural populations and, as in this example, can be detected most easily by molecular methods. In the DNA of a gene, each nucleotide can be A, T, G, or C, so a gene of n nucleotides can theoretically mutate at any of the positions to any of the three other nucleotides. The number of possible single-nucleotide differences in a gene of length n is, therefore, 3 × n. If n = 5000, for example, there are potentially 15,000 alleles (not counting any of the possibilities with more than one nucleotide substitution). Most of the potential alleles do not actually exist at any one time. Some are absent because they did not occur, others did occur but were eliminated by chance or because they were harmful, and still others are present but at such a low frequency that they remain undetected. Nevertheless, at the level of DNA sequence, most genes in most natural populations have multiple alleles, all of which can be considered “wild type.” Multiple wild-type alleles are useful in such applications as DNA typing because two unrelated people are unlikely to have the same genotype, especially if several different loci, each with multiple

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

alleles, are examined. Many harmful mutations also exist in multiple forms. Recall from the Genes, Genomes, and Genetic Analysis chapter that more than 400 mutant forms of the phenylalanine hydroxylase gene have been identified in patients with phenylketonuria. The alleles A1–A6 in Figure 3.16 also illustrate

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

that although a population of organisms may contain any number of alleles, any particular organism or cell may carry no more than two, and any gamete may carry no more than one. In some cases, the multiple alleles of a gene exist merely by chance and reflect the history of mutations that have taken place in the population and the dissemination of these mutations among population subgroups by migration and interbreeding. In other cases, biological mechanisms favor the maintenance of a large number of alleles. For example, genes that control self-sterility in certain flowering plants can have large numbers of allelic types. This type of self-sterility is found in species of red clover that grow wild in many pastures. The self-sterility genes prevent selffertilization because a pollen grain can undergo pollen tube growth and fertilization only if it contains a self-sterility allele different from either of the alleles present in the flower on which it lands. In other words, a pollen grain containing an allele already present in a flower will not function on that flower. Because all pollen grains produced by a plant must contain one of the self-sterility alleles present in the plant, pollen cannot function on the same plant that produced it, and self-fertilization cannot take place. Under these conditions, any plant with a new allele can fertilize all flowers except those on the same plant. Through evolution, populations of red clover have accumulated hundreds of alleles of the self-sterility gene, many of which have been isolated and their DNA sequences determined. Many of the alleles differ at multiple nucleotide sites, which implies that the alleles in the population are very old.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Human ABO Blood Groups In a multiple allelic series, there may be different dominance relationships between different pairs of alleles. An example is found in the human ABO blood groups, which are determined by three alleles denoted IA, IB, and IO. (Actually, there are two slightly different variants of the IA allele.) The blood group of any person may be A, B, AB, or O, depending on the type of polysaccharide (polymer of sugars) present on the surface of red blood cells. One of two different polysaccharides, A or B, can be formed from a precursor molecule that is modified by the enzyme product of the IA or the IB allele. The gene product is a glycosyl transferase enzyme that attaches a sugar unit to the precursor (FIGURE 3.18). The IA and IB alleles encode different forms of the enzyme with replacements at four amino acid sites; these alter the substrate specificity so that each enzyme attaches a different sugar. People of genotype IAIA produce red blood cells having only the A polysaccharide and are said to have blood type A. Those of genotype IBIB have red blood cells with only the B polysaccharide and have blood type B. Heterozygous IAIB people

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

have red cells with both the A and the B polysaccharides and have blood type AB. The third allele, IO, encodes an enzymatically inactive protein that leaves the precursor unchanged; neither the A nor the B type of polysaccharide is produced. Thus, homozygous IOIO persons lack both the A and the B polysaccharides and are said to have blood type O.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

FIGURE 3.18 The ABO antigens on the surface of human red blood cells are carbohydrates.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In this multiple allelic series, the IA and IB alleles are codominant: The heterozygous genotype has the characteristics of both homozygous genotypes—the presence of both the A and B carbohydrates on the red blood cells. In contrast, the IO allele is recessive to both IA and IB. Hence, heterozygous IAIO genotypes produce the A polysaccharide and have blood type A, and heterozygous IBIO genotypes produce the B polysaccharide and have blood type B. The genotypes and phenotypes of the ABO blood group system are summarized in the first three columns of TABLE 3.4.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

TABLE 3.4 Genetic control of the human ABO blood group Genotype

Antigens

ABO

Antibodies

Blood

Blood types

present

blood

present in

types that

that can

on red

group

blood fluid

can be

accept

blood

phenotype

tolerated in

blood for

transfusion

transfusion

cells IAIA

A

Type A

Anti-B

A & O

A & AB

IAIO

A

Type A

Anti-B

A & O

A & AB

IBIB

B

Type B

Anti-A

B & O

B & AB

IBIO

B

Type B

Anti-A

B & O

B & AB

IAIB

A & B

Type AB

Neither

A, B, AB, &

AB only

anti-A nor

O

anti-B IOIO

Neither A

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

nor B

Type O

Anti-A & anti-B

O only

A, B, AB, & O

ABO blood groups are important in medicine because of the need for blood transfusions. A crucial feature of the ABO system is that most human blood contains antibodies to either the A or B polysaccharide. An antibody is a protein made by the immune system in response to a stimulating molecule called an antigen and is capable of binding to the antigen. An antibody is usually specific in that it recognizes only one antigen. Some antibodies combine with antigen and form large molecular aggregates that may precipitate. Antibodies act to defend the body against invading viruses and bacteria. Although antibodies do not normally form without prior stimulation by the antigen, people capable of producing anti-A and

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

anti-B antibodies do produce them. Production of these antibodies may be stimulated by antigens that are similar to polysaccharides A and B and that are present on the surfaces of many common bacteria. However, a mechanism called tolerance prevents an organism from producing antibodies against its own antigens. This mechanism ensures that A antigen or B antigen elicits antibody production only in people whose own red blood cells do not contain A or B, respectively. The end result: People of blood type O make both anti-A and anti-B antibodies, those of blood type A make anti-B antibodies, those of blood type B make anti-A antibodies, and those of blood type AB make neither type of antibody.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.19 is an example of how antibodies can be used to determine blood type.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.19 The reaction between the indicated blood types and anti-A (left column) and anti-B (right column). © Chatsikan Tawanthaisong/Shutterstock.

The antibodies found in the blood fluid of people with each of the ABO blood types are shown in the fourth column in Table 3.4. The clinical significance of the ABO blood groups is that transfusion of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

blood containing A or B red-cell antigens into persons who make antibodies against them results in an agglutination reaction in which the donor red blood cells are clumped. In this reaction, the anti-A antibody agglutinates red blood cells of either blood type A or blood type AB, because both carry the A antigen. Similarly, anti-B antibody agglutinates red blood cells of either blood type B or blood type AB. When the blood cells agglutinate, many blood vessels are blocked, and the recipient of the transfusion goes into shock and may die. Incompatibility in the other direction, in which the donor blood contains antibodies against the recipient's red blood cells, is usually acceptable because the donor's antibodies are diluted so rapidly that clumping is avoided. The types of compatible blood transfusions are shown in the last two columns of Table 3.4. Note that a person of blood type AB can receive blood from a person of any other ABO type; type AB is called a universal recipient. Conversely, a person of blood type O can donate blood to a person of any ABO type; type O is called a universal donor.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Epistasis In the Genes, Genomes, and Genetic Analysis chapter, we saw how the products of several genes may be necessary to carry out all the steps in a biochemical pathway. In genetic crosses in which two mutations that affect different steps in a single pathway are both segregating, the typical F2 ratio of 9 : 3 : 3 : 1 is not observed. Gene interaction that perturbs the normal Mendelian ratios is known as epistasis. One type of epistasis is illustrated by the interaction of the C, c and P, p allele pairs affecting flower coloration in peas. These genes encode enzymes in the biochemical pathway for the synthesis of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

anthocyanin pigment, and the production of anthocyanin requires the presence of at least one wild-type dominant allele of each gene. We can designate this genotype as

Suppose we do a cross between CC pp and cc PP, as in FIGURE 3.20A. The phenotype of the flowers in the F1 generation is the wild-type purple. Why? Because the C allele is dominant to c and the P allele is dominant to p, the F1 plant is a double heterozygote of genotype Cc Pp and, therefore, has purple flowers. The result is nevertheless strange, because the original cross involved two homozygous recessive mutants, each of which had white flowers. Once we have been told that the mutant c and the mutant p alleles are in different genes, the finding of wild-type flowers in the F1

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

generation is logical.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.



FIGURE 3.20 Epistasis in the determination of flower color in peas. Formation of the purple pigment requires the dominant allele of both the C and P genes. (A) Parental and F1 crosses. (B) Forked line diagram showing that with this type of epistasis, the dihybrid F2

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

ratio is modified to 9 purple : 7 white.

But what if we did not know whether the mutant alleles were in different genes? What if both mutants were picked up in a mutant screen, or discovered in different laboratories, and we did not know whether the c allele and the p allele were alleles of the same gene or alleles of different genes? The answer is that the phenotype of the F1 progeny in a cross like that in Figure 3.20 tells you. On the one hand, if the phenotype of the F1 progeny is wild type, as in Figure 3.20, this observation tells you that c and p are alleles of different genes. On the other hand, if the phenotype of the F1 progeny is mutant (in this case, a plant with white flowers),

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

this result tells you that c and p are alleles of the same gene. In the latter case, it would be best to change their names in such a way as to signify their allelism, such as c = c1 and p = c2, because then the genotype of the F1 plant would be written as c1/c2 and it would be obvious from the genotype symbol that the phenotype is mutant, because each allele of the gene is mutant. This discussion should remind you of the Beadle–Tatum experiments—in particular, their use of the complementation test to determine whether two nutritional mutants in Neurospora were, or were not, mutants of the same gene. The principle in Figure 3.20 is exactly the same. The difference is that Beadle and Tatum used Neurospora heterokaryons, in which mutant alleles are brought together by forming hybrid filaments with two types of haploid mutant nuclei, whereas in diploid organisms, like the peas in Figure 3.20, the hybrid nuclei are created by crossing two homozygous recessive mutants. The main point is that complementation tests are also used in sexual diploid organisms to identify which recessive mutations are alleles and which are not, and the phenotype of the F1 generation in Figure 3.20 illustrates the underlying principle. Whereas the F1 generation in Figure 3.20A illustrates Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

complementation, the F2 generation illustrates one type of epistasis. Suppose that plants of the F1 generation are selffertilized (indicated by the encircled cross sign), and assume that the (C, c) and (P, p) allele pairs undergo independent assortment. The forked line diagram in FIGURE 3.20B gives the expected phenotypes of the F2 generation. Because only the C— P— progeny have purple flowers, the ratio of purple flowers to white flowers in the F2 generation is 9 : 7. The epistasis does not change the result of independent segregation; it merely conceals the fact

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

that the underlying ratio of the genotypes C— P— : C— pp : cc P — : cc pp is 9 : 3 : 3 : 1. For a trait determined by the interaction of two genes, each with a dominant allele, there are only a limited number of ways in which the 9 : 3 : 3 : 1 dihybrid ratio can be modified. The possibilities are illustrated in FIGURE 3.21. In the absence of epistasis, the F2 ratio of phenotypes is 9 : 3 : 3 : 1. In each row, different colors indicate different phenotypes. For example, in the modified ratio at the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

bottom, the phenotypes of the “3 : 3 : 1” classes are indistinguishable, resulting in a 9 : 7 ratio. This ratio is observed in the segregation of the C, c and P, p alleles in Figure 3.20, with its 9 : 7 ratio of purple to white flowers. Taking all the possible modified ratios in Figure 3.21 together, there are nine possible dihybrid ratios when both genes show complete dominance. Examples of each of the modified ratios are known. The types of epistasis that result in these modified ratios are illustrated in the following examples, which are taken from a variety of organisms. Other examples can be found in the problems at the end of the chapter.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

FIGURE 3.21 Modified F2 dihybrid ratios. In each row, different colors indicate different phenotypes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

9 : 7 is observed when a homozygous recessive mutation in either or both of two different genes results in the same mutant phenotype, as in Figure 3.20. 12 : 3 : 1 results when the presence of a dominant allele at one locus masks the genotype at a different locus, such as the A— genotype rendering the B— and bb genotypes indistinguishable. For example, in genetic study of the color of the hull in oat seeds, a variety with white hulls was crossed with a variety with black hulls. The F1 hybrid seeds had black hulls. Among 560 progeny in the F2 generation, the hull phenotypes observed were 418 black, 106 gray, and 36 white. The ratio of phenotypes is 11.6 : 3.9 : 1, or very nearly 12 : 3 : 1. A genetic hypothesis to explain these results is that the black-hull phenotype is due to the presence of a dominant allele A and the gray-hull phenotype is due to another dominant allele

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

B whose effect is apparent only in aa genotypes. On the basis of this hypothesis, the original varieties had genotypes aa bb (white) and AA BB (black). The F1 generation has genotype Aa Bb (black). If the A, a allele pair and the B, b allele pair undergo independent assortment, then the F2 generation is expected to have the genotypic and phenotypic composition 9/16 A— B— (black hull), 3/16 A— bb (black hull), 3/16 aa B— (gray hull), 1/16 aa bb (white hull), or 12 : 3 : 1. 13 : 3 is illustrated by the difference between White Leghorn chickens (genotype CC II) and White Wyandotte chickens (genotype cc ii). Both breeds have white feathers because the C allele is necessary for colored feathers, but the I allele in White Leghorns is a dominant inhibitor of feather coloration. The F1 generation of a dihybrid cross between these breeds has the genotype Cc Ii, which is expressed as white feathers because of the inhibitory effects of the I allele. In the F2

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

generation, only the C— ii genotype has colored feathers, so there is a 13 : 3 ratio of white : colored. 9 : 4 : 3 is observed when homozygosity for a recessive allele with respect to one gene masks the expression of the genotype of a different gene. For example, if the aa genotype has the same phenotype regardless of whether the genotype is B— or bb, then the 9 : 4 : 3 ratio results. As an example, in mice the grayish “agouti” coat color results from a horizontal band of yellow pigment just beneath the tip of each hair. The agouti pattern is due to a dominant allele A, and in aa animals the coat color is black. A second dominant allele, C, is necessary for the formation of hair pigments of any kind, and cc animals are albino (white). In a cross of AA CC (agouti) × aa cc (albino), the F1 progeny are Aa Cc and phenotypically agouti. Crosses between F1 males and females produce F2 progeny in the proportions 9/16 A— C— (agouti), 3/16 A— cc (albino), 3/16

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

aa C— (black), 1/16 aa cc (albino), or 9 agouti : 4 albino : 3 black. 9 : 6 : 1 implies that homozygosity for either of two recessive alleles yields the same phenotype but that the phenotype of the double homozygote is different. In Duroc-Jersey pigs, red coat color requires the presence of two dominant alleles R and S. Pigs of genotype R— ss and rr S— have sandy-colored coats, and rr ss pigs are white. The F2 ratio is therefore 9/16 R— S— (red), 3/16 R— ss (sandy), 3/16 rr S— (sandy), 1/16 rr ss (white), or 9 red : 6 sandy : 1 white.

SUMMING UP ■ Incomplete dominance exists when heterozygotes for a gene can be phenotypically distinguished from both homozygotes. ■ In a population, more than two alleles for a population can exist, and dominance relationships among pairs of them can differ. ■ Epistatic interactions among genes results in perturbation of normal Mendelian ratios.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Epistatic genes often encode different enzymes that are involved in a particular biochemical pathway.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.6 Probability in Genetic Analysis Mendel's laws of genetic transmission are fundamentally laws of chance (probability). Mendel surpassed any of his contemporaries in understanding that his principles of inheritance accounted for the different types of progeny he observed as well as for the ratios in which they were found. No discoveries in genetics made since Mendel's time have undermined the fundamental role of chance in

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

heredity that he was the first to recognize. To fully understand Mendelian genetics, then, we need to understand the elementary principles of probability. Every problem in probability begins with an experiment, which may be real or imaginary. In genetics, the experiment is typically a cross. Associated with the experiment is a set of possible outcomes of the experiment. In genetics, the possible outcomes are typically genotypes or phenotypes. The possible outcomes are called elementary outcomes. They are elementary outcomes in the sense that none of them can be reduced to a combination of the others. For example, in the case of genetic crosses, elementary outcomes are the different phenotypes that are possible in the progeny. In our applications of probability, the number of elementary outcomes is often relatively small, or in any case can be enumerated. The principles of probability also can deal with conceptual experiments in which there are an infinite number of elementary outcomes, but this requires some technicalities that are beyond the scope of this text. Each elementary outcome is assigned a probability that is proportional to its likelihood of occurrence. In principle, the probabilities can be assigned arbitrarily. There are only two rules. First, the probability of each elementary outcome must be a nonnegative number between 0 and 1, and may actually equal 0 or 1.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

An elementary outcome with a probability of 0 cannot occur, and one with a probability of 1 must occur. Second, the sum of the probabilities of all the elementary outcomes must equal 1. This rule assures that at least one of each elementary outcome must occur. These two rules also handle an annoying question often asked in regard to a coin toss: What happens if it lands on its edge? The answer is that this elementary outcome is assigned a probability of 0, so we need not bother with it. Here it will be helpful to consider a specific example. Consider the conceptual experiment of self-fertilization of the F1 progeny of a

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

cross between pea plants homozygous for the round W and yellow G alleles and plants homozygous for w and g. Recall that the W, w and G, g allele pairs undergo independent assortment, so there are 16 elementary outcomes. The genotypes are shown in the Punnett square in Figure 3.9, but are illustrated in a different manner in FIGURE 3.22A. Each of the elementary outcomes is equally likely, so the probability of each elementary outcome is assigned a value of 1/16. Note that the progeny genotype Ww Gg is listed four times, reflecting the fact that there are four possible combinations of parental gametes that can yield the progeny genotype Ww Gg.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.22 (A) Sample space for the possible progeny in a cross for the allele pairs W, w for round versus wrinkled seeds and G, g for yellow versus green seeds. (B) The event A includes all genotypes whose phenotype is wrinkled. (C) The event B includes all genotypes whose phenotype is yellow. (D) The union and intersection (yellow) of A and B.

The enumeration of the elementary outcomes and their probabilities constitute what is often called the sample space of the probability problem. For the progeny from self-fertilization of Ww Gg, the sample space is shown in Figure 3.22A. This is the sample space in which all probability considerations regarding this conceptual experiment take place.

Elementary Outcomes and Events

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Any combination of elementary outcomes constitutes an event in the sample space. In FIGURE 3.22B, the circle labeled A includes four elementary outcomes defining the event A as “the offspring genotype is ww.” The event A could be defined equivalently as “the offspring phenotype is wrinkled.” The alternative ways of defining A in words show how subsets of elementary outcomes can relate genotypes and phenotypes. Corresponding to every event is a probability of that event, in this case symbolized Pr{A}. A fundamental principle of probability is that: The probability of any event equals the sum of the probabilities of all the elementary outcomes included in the event.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In the example in Figure 3.22B, Pr{A} = 4/16 because the event A consists of four elementary outcomes (possible progeny genotypes), each of which has probability of 1/16. An event may include no elementary outcomes, in which case its probability is 0; or it may include all elementary outcomes, in which case its probability is 1. The elementary outcomes outside the circle defining event A in Figure 3.22B define another event that we have denoted A′. This event consists of all elementary outcomes not present in A; in other words, it consists of all progeny whose genotype is not ww, or equivalently it consists of progeny whose phenotype is round. The event A′ is called the complement of A, and it may also be denoted as AC, A, or not-A. The probability of A′ again equals the sum of the probabilities of the elementary outcomes that constitute A′, so in this case Pr{A′} = 12/16.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Events also can be composed of other events. To see how, consider the event denoted as B in FIGURE 3.22C. Event B is defined as all progeny from the cross whose genotype is gg, which could also be defined as all progeny whose seeds are green. Because B includes four elementary outcomes, its probability is Pr{B} = 4/16. Again, there is a complementary event B′ defined as all elementary outcomes not included in B; to define B′ in another way, all progeny whose genotype is either GG or Gg. There are 12 elementary outcomes in B′, so Pr{B′} = 12/16.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Look again at Figures 3.22B and C and consider the event consisting of elementary outcomes that are included in A, included in B, or both. This event is called the union of A and B. We will denote the union of A and B as A + B, but in many probability textbooks you may find it symbolized as A B, where the symbol is pronounced “cup.” The event A + B consists of seven elementary outcomes in which the progeny genotype is either ww, gg, or both, so Pr{A + B} = 7/16. Another important way in which two events A and B can be combined is called the intersection of A and B. It consists of all the elementary outcomes that are included in both A and B. An example is shown in Figure 3.22D, where the intersection of A and B is shaded yellow. We will denote the intersection of A and B as AB, but in probability textbooks the intersection is often represented as A∩B, where the symbol ∩ is pronounced “cap.”

Probability of the Union of Events As with all events, the probability of the union of events A + B equals the sum of the probabilities of the elementary outcomes that are included in A or B or both. Likewise, the probability of the intersection of events AB equals the sum of the probabilities of the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

elementary outcomes that are included in both A and B. In general, an equation for the probability of the union of A and B is:

The reason for subtracting Pr{AB} becomes evident when we look at Figure 3.22D. Because the progeny genotype ww gg is included in both A and B, when the probabilities of the elementary outcomes in A are added to those in B, the genotype ww gg is included twice. To correct for the overcounting, the probability of this outcome must be subtracted from the total. Because the elementary outcomes counted twice are exactly those that are present in both A and B, they constitute the intersection of A and B. Hence, Pr{AB} is the quantity that is subtracted in Equation 1.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

An important special case of Equation 1 arises when A and B do not overlap—that is, when they have no elementary outcomes in common. In this case, A and B are said to be mutually exclusive (or disjoint), and Pr{AB} = 0. Therefore, when A and B are mutually exclusive, Equation 1 becomes:

This equation is sometimes called the addition rule for mutually exclusive events. Events are mutually exclusive when the occurrence of one event excludes the occurrence of the other. In other words, mutually exclusive events are mutually incompatible in the sense that no elementary outcome can be present in both. An example, again based on the experiment of crossing Ww Gg × Ww Gg, is shown in FIGURE 3.23. Here we have defined events A* and B* in such a way that they are modified versions of A and B that do not overlap. In words, the event A* may be defined as “the progeny phenotype is wrinkled but not green,” and the event B*

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

may be defined as “the progeny phenotype is green but not wrinkled.” Defined in this way, A* and B* are mutually exclusive, and therefore Equation 2 has the implication that Pr{A* + B*} = Pr{A*} + Pr{B*} = 6⁄16.

FIGURE 3.23 The events A* and B* are defined in such a way that each excludes genotypes whose phenotype is wrinkled and green. A* and B* do not overlap and, therefore, are mutually exclusive.

Probability of the Intersection of Events

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Returning again to events A and B in Figure 3.22D, consider the probability of the intersection AB. This event consists of all elementary outcomes that are included in both A and B, which in this example consists of one genotype only—namely, ww gg. The probability of AB is therefore Pr{AB} = 1/16. This is a special case in which the events are independent, which means that knowing whether the event A occurs tells you nothing about whether the event B occurs. The probability of the joint occurrence of independent events has the special property that it is proportional to the product of the probabilities of the individual events. In the example in Figure 3.22D, we saw that Pr{A} = 4/16 and that Pr{B} = 4/16; in this case, then, Pr{AB} = 4/16 × 4/16 = 16/256 = 1/16.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

The events A and B in this example illustrate an important general principle: When the events A and B are independent, their probability is given by

This equation is sometimes called the multiplication rule for independent events. The choice of terms for “independent events”

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

and “independent assortment” is not fortuitous, because independent assortment means that knowing the genotype of an offspring for the W, w allele pair tells you nothing about the genotype for the G, g allele pair. This principle is illustrated in FIGURE 3.24A.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.24 In genetics, two important types of independence are (A) independent segregation of alleles that show independent assortment and (B) independent fertilizations resulting in successive offspring.

Another situation in genetics in which independence is the rule is shown in FIGURE 3.24B, which deals with successive offspring from a cross. Successive offspring are independent because the genotypes of early progeny have no influence on the probabilities in later progeny. The independence of successive offspring contradicts the widespread belief that in each human family, the ratio of girls to boys must “even out” at approximately 1:1. According to this reasoning, a family with four girls would be more likely to have a boy the next time around. In reality, this belief is supported neither by theory nor by actual data. The data indicate that human families are equally likely to have a girl or a boy on any

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

birth, irrespective of the sex distribution in previous births. Although statistics guarantees that the sex ratio will balance out when averaged over a very large number of cases, this does not imply that it will equalize in any individual case. To make this principle more concrete, consider that among families in which there are five children, those consisting of five boys balance those consisting of five girls, yielding an overall sex ratio of 1 : 1. Nevertheless, both of these sex ratios are unusual.

SUMMING UP ■ An experiment has elementary outcomes, each one of which has some probability of occurring. ■ An event is some combination of elementary outcomes, and its probability is the sum of the probabilities of those events occurring. ■ The probability of either of two mutually exclusive events occurring is equal to the sum of the probabilities of those two events.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The probability of two independent events occurring simultaneously is the product of the probabilities of those two events.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.7 Conditional Probability and Pedigrees Let us now return to pedigree analysis and think about how probabilistic thinking can be applied to it. The analysis of human pedigrees creates many situations in which information is known about a person that can affect the assessment of the likelihood that the individual has a particular genotype or phenotype. Think back to Huntington disease, which is caused by an autosomal dominant allele. The incidence of this disease worldwide is approximately 2.7 cases per 100,000 individuals, meaning that in the absence of any other information, the probability of a particular individual contracting the disease is 2.7/100000, or 2.7 × 10−5. But what if we have some basic pedigree information? Consider the following possibilities:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1. Neither parent of the individual develops Huntington disease. In that case, ignoring for now the possibility that a new mutation occurs, the probability of the individual developing the disease is zero. 2. One of the individual's parents does, in fact, develop Huntington disease. If so, the probability that the offspring individual received the disease allele from the parent, and thus that she or he will develop the disease, is now 1/2. These are examples of conditional probabilities. A conditional probability can be thought of as the probability that a particular event A occurs given the occurrence of event B. For example, we could reframe case 2 to be “Given that an individual's parent develops Huntington disease (event A), what is the probability that she or he will develop the disease (event B) as well?” In formal probability notation,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

This example is relatively straightforward, in that Huntington disease alleles are dominant and for the most part completely penetrant. But what if those conditions do not hold for a gene or phenotype of interest? The following two problems provide hypothetical examples.

Problem 1. In the pedigree in FIGURE 3.25, the female I-1 is heterozygous for a mutant gene M that increases the risk of a certain disease from 5 percent observed among mm genotypes to 50 percent in Mm. The M allele is associated with the larger DNA band in the gel (the band nearer the top). The male I-2 is homozygous mm for the nonmutant form of the gene, and the m allele is associated with the smaller DNA band (the one nearer the bottom). Their daughter II-1 has the disease. What is the probability that II-1 has the genotype Mm?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 3.25 The rare M allele in this pedigree, associated with the DNA nearer the top of the gel, increases the risk of a disease from 5 percent in homozygous mm genotypes to 50 percent in heterozygous Mm genotypes. A typical problem in conditional probability is to calculate the probability that the daughter II-1 has genotype Mm, given that she is affected.

To get at this problem, let's imagine there are 100 pedigrees like Figure 3.25 instead of just one. Simple Mendelian inference tells us that we would expect 50 of the progeny in them to be Mm and 50 to be mm. If we then consider what we know about the expected frequency of the trait in each of the two genotypes, we would

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

expect 5 of the mm individuals and 50 of the Mm ones to be affected (a total of 55 people). Thus, the probability that an affected individual is Mm is simply the fraction of all of the affected individuals who are MM—that is, 50/55, or 0.91. This probability is far greater than the 1/2 = 0.50 expected from Mendelian segregation, and the discrepancy shows that using conditional probability based on prior knowledge that the daughter is affected has a major impact on the assessment of how likely it is that she has genotype Mm. The following problem also makes use of conditional probability but in a somewhat different way.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Problem 2. Consider again the M mutation in Figure 3.25. Imagine a human population in which the proportion of people of genotype Mm is 1 percent and in which homozygous MM individuals are so rare that they can be ignored. Given that a person in this population is affected with the disease, what is the probability that the affected individual has genotype Mm? In the person in question, the disease condition is already known to exist, and given this condition we need to deduce the probability that the affected individual's genotype is Mm. We will use the same approach, but instead of having prior knowledge based on a pedigree, we know only the frequency of the allele in the population. To do this, imagine a sample of 2000 people from this population. Among these, 20 (that is, 1 percent) have genotype Mm, of whom 10 (that is, 50 percent) will be affected. The remaining 1980 people have genotype mm, of whom 99 (that is, 5 percent) will be affected. Among the people who are affected, the probability of genotype Mm is 10/(10 + 99) = 10/109 = 0.092. Note that, even though genotype Mm greatly

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

increases the chance of the disease, the great majority of affected people (99 out of 109) nevertheless have genotype mm. In Problem 1, the probability that an affected individual has genotype Mm is 0.91, whereas in Problem 2 it is 0.092. There is about a tenfold difference in these probabilities. Where does the difference come from? It comes from the vastly different prior information that we apply to the problem. In Problem 1, we know that the mother has genotype Mm, so the probability that an offspring has genotype Mm is known in advance to be 0.5 (this is called the prior probability). In Problem 2, we consider the population as a whole, and in this case the prior probability of Mm is said to be 0.01 (1 percent). Despite the very different prior probabilities, however, we were able to navigate through the problem and get the right answers. The next section discusses reasoning based on conditional probability in more formal terms.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Bayes' Theorem In the previous two examples, we used what is called the method of natural frequencies. That is, we generated hypothetical populations in which we then calculated posterior probabilities— that is, the probability of affected individuals having the Mm genotype based on their expected frequencies in those populations. In fact, it is possible to go directly from prior information such as we were given in these problems to posterior probabilities, using conditional probabilities. This insight is credited to Thomas Bayes (1702–1761), a Presbyterian minister in Tunbridge Wells, England. His ideas were summarized in his Essay Towards Solving a Problem in the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Doctrine of Chances, published in 1763, two years after his death. To understand his idea, we return to the concept of conditional probability, which can be formally defined as the probability that both A and B occur together (symbolized Pr{AB}), divided by the probability of event B (symbolized Pr{B}), provided that Pr{B} is not equal to 0. Written as an equation

where Pr{B|A} is expressed in words as “the probability of B given A.”

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In Problem 1, for example, suppose we let A be the event that individual II-1 in Figure 3.25 is affected and B be the event that individual II-1 has genotype Mm. We are given the prior information that the probability that II-1 is affected, given her genotype of Mm, equals 0.50; hence, Pr{A|B} = 0.50. However, what the problems asks us to calculate is the probability that II-1 has genotype Mm, given that she is affected; this probability is symbolized as Pr{B|A}, and it is not the same as Pr{A|B}. How can we apply what we know to calculate Pr{B|A} using Equation 4? We start with the denominator. It is the probability that A occurs together with B or that A occurs together with B′; this sum must equal the probability that A occurs, since if A occurs it must occur in combination either with B or with B′. Thus,

Now we turn to the numerator in Equation 4. It is the probability that A occurs, given that B has occurred (Pr{A|B}, times the probability that B occurs in the first place (Pr{B}). We can thus rewrite equation 4 as follows:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Based on our prior information about the trait and our knowledge of the pedigree, we can solve the problem:

In fact, Equation 5 is Bayes' theorem. To further illustrate its usefulness, consider Problem 2. In this case, we use the prior information that Pr{an individual is affected | genotype Mm} = 0.50 and Pr{an individual is affected | genotype mm} = 0.05, which tells us that we should choose A = “an individual is affected” and B = “the individual has genotype Mm.” In this problem, we are not told the parental genotypes but rather given the population frequencies of Mm as Pr{B} = 0.01 and mm as Pr{B′} = 0.99. These considerations provide the factors needed for the right-hand side of Bayes' theorem (Equation 5):

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

This result matches the answer calculated earlier using natural frequencies.

Problem 3. As one more example of the use of conditional probability in genetic calculations, we consider a problem that uses only the denominator in Equation 5. Consider the human population in Problem 2 consisting of 1 percent of individuals of genotype Mm and 99 percent mm. If a person from this population is chosen at random, what is the probability that the person is affected? We have already calculated that, in a sample of 2000 people in this

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

population, 10 with genotype Mm will be affected and 99 of genotype mm will be affected. Hence the overall probability of an affected individual is 109/2000 = 0.0545. Alternatively, we can use the denominator of Equation 5, with A = “an individual is affected” and B = “the individual has genotype Mm,” so that Pr{A|B} = 0.50, Pr{A|B′} = 0.05, Pr{B} = 0.01, and Pr{B′} = 0.99. Then,

Note that 0.0545 is not much greater than 0.05, which is the probability that an individual is affected given that the genotype is mm. The reason that the overall risk of being affected is so close to 0.05 is that Mm genotypes make up such a small proportion of the overall population.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

For many students, the main issue with Bayes' theorem (Equation 5) is that it looks so intimidating. A question often asked is, “How do I know what to choose as event A and what to choose as event B?” The choice of A and B is determined by choosing them so that Pr{A|B} is prior information and Pr{B|A} is the probability you want to know. In solving problems in conditional probability, which method should you use—natural frequencies or Bayes' theorem? You should use whichever method suits you best. Some students prefer the informal style of natural frequencies, whereas others like the more formal approach of Bayes' theorem. Whichever method you decide you like better, you should try to understand both. This is an extremely important concept in terms of genetic risk assessment, especially when the penetrance of a trait is less than 100 percent or the trait is not a simple Mendelian one.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

SUMMING UP ■ Determination of conditional probabilities is a critical part of pedigree analysis. ■ Conditional probabilities can be calculated based on the expected frequencies of different outcomes in large hypothetical populations. ■ Bayes' theorem provides a formal means whereby conditional probabilities can be calculated based on prior information.

CHAPTER SUMMARY ■ Inherited traits are determined by the genes present in the reproductive cells united in fertilization. ■ Genes are usually inherited in pairs: one from the mother and one from the father. ■ The genes in a pair may differ in DNA sequence and in their effect on the expression of a particular inherited trait.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The maternally and paternally inherited genes are not changed by being together in the same organism. ■ In the formation of reproductive cells, the paired genes separate again into different cells. ■ Random combinations of reproductive cells containing different genes result in Mendel's ratios of traits appearing among the progeny. ■ The ratios actually observed for any trait are determined by the types of dominance and gene interaction. ■ The laws of probability provide the basis for making predictions based on genetic hypotheses.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

REVIEW THE BASICS ■ What is the principle of segregation, and how is this principle demonstrated in the results of a single-gene (monohybrid) cross? ■ What is the principle of independent assortment, and how is this principle demonstrated in the results of a two-gene (dihybrid) cross? ■ Explain why random union of male and female gametes is necessary for Mendelian segregation and independent assortment to be observed in the progeny of a cross. ■ When two pairs of alleles show independent assortment, under which conditions will a 9 : 3 : 3 : 1 ratio of phenotypes in the F2 generation not be observed? ■ Explain the following statement: “Among the F2 progeny of a dihybrid cross, the ratio of genotypes is 1 : 2 : 1, but among the progeny that express the dominant phenotype, the ratio of genotypes is 1 : 2.” ■ What are the principal features of human pedigrees in which a rare dominant allele is segregating? In which a rare recessive allele is segregating?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ What is a mutant screen and how is it used in genetic analysis? ■ Explain the following statement: “In genetics, a gene is identified experimentally by a set of mutant alleles that fail to show complementation.” What is complementation? How does a complementation test enable a geneticist to determine whether two different mutations are or are not mutations in the same gene? ■ What does it mean to say that epistasis results in a “modified dihybrid F2 ratio”? Give two examples of a modified dihybrid F2 ratio, and explain the gene interactions that result in the modified ratio.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

■ What is the difference between mutually exclusive events and independent events? How are the probabilities of these two types of events combined? Give two examples of genetic events that are mutually exclusive and two examples of genetic events that are independent. ■ What is a conditional probability? How does Bayes' theorem allow us to calculate one?

GUIDE TO PROBLEM SOLVING PROBLEM 1 Complete the table by inserting 0, 1/4, 1/2, or 1 for the probability of each genotype of the progeny from each type of mating. For which mating are the parents identical in genotype but the progeny as variable in the genotype as they can be for a single locus? For which mating are the parents as different as they can be for a single locus but the progeny identical to each other and different from either parent?

Progeny genotypes Mating

AA

Aa

aa

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

AA × AA AA × Aa AA × aa Aa × Aa Aa × aa aa × aa

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

ANSWER This kind of table is fundamental to being able to solve almost any type of quantitative problem in transmission genetics.

Progeny genotypes Mating

AA

Aa

aa

AA × AA

1

0

0

AA × Aa

1/2

1/2

0

AA × aa

0

1

0

Aa × Aa

1/4

1/2

1/4

Aa × aa

0

1/2

1/2

aa × aa

0

0

1

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The mating for which the parents are identical in genotype, but the progeny are as variable in genotype as they can be for a single locus, is Aa × Aa. The mating for which the parents are as different as they can be for a single locus, but the progeny identical to each other and different from either parent, is AA × aa. PROBLEM 2 In domesticated chickens, a dominant allele C is required for colored feathers, but a dominant allele I of an unlinked gene is an inhibitor of color that overrides the effects of C. White Leghorns have genotype CC II, whereas White Wyandottes have genotype cc ii. Both breeds are white, but for different reasons. In the F2 generation of a cross between White Leghorns and White Wyandottes: (a) What is the phenotypic ratio of white : colored? (b) Among F2 chicks that are white, what is the proportion of Cc Ii?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

ANSWER (a) The initial cross of CC II × cc ii will yield F1 chickens with genotype Cc Ii. Because the genes are not linked, the offspring in the F2 generation will have a ratio of the genotypes 9 C— I— : 3 C— ii : 3 cc I— : 1 cc ii. Only the chickens with genotype C — ii will be colored, because C is required for colored feathers but I inhibits the effect of C. This type of interaction of genes, or epistasis, will lead to a phenotypic ratio of 13 white : 3 colored. (b) In any dihybrid cross with unlinked genes, the probability that an F2 progeny is heterozygous for both genes equals (1/2) × (1/2) = 1/4. Because only 13/16 of the chicks in this example are white, the proportion of Cc Ii among the white chicks is (1/4)/(13/16) = 4/13. You can also get this result by counting squares in the Punnett square for a dihybrid cross: A total of 13 genotypes yield white chicks, and among these four are heterozygous for both genes. PROBLEM 3 In Duroc-Jersey pigs, animals with genotype R— S — are red, those with genotype R— ss or rr S— are sandy colored, and those with the genotype rr ss are white. The genes show independent assortment. In the F2 generation of a cross

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

between the genotypes RR ss and rr SS: (a) What is the phenotypic ratio of red : nonred? (b) Among F2 piglets that are red, what is the proportion of Rr Ss? ANSWER (a) The initial cross RR ss × rr SS will yield F1 pigs with genotype Rr Ss. The progeny of the F2 generation will have the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

genotypic ratio 9 R— S— : 3 R— ss : 3 rr S— : 1 rr ss. Only the pigs with genotype R— S— will be red, while all the others will be nonred (some sandy and some white, in the ratio of 6 : 1). This type of interaction of two genes (epistasis) will lead to a phenotypic ratio of 9 red : 7 nonred. (b) 4/9. The proportion of pigs with the genotype Rr Ss is the frequency of those heterozygous for both genes, which equals 1/2 × 1/2 = 1/4. Because only 9/16 of the pigs are red, the proportion of Rr Ss among them is (1/2 × 1/2)/(9/16) = 4/9. You can also get this answer by counting squares in the Punnett square for a dihybrid cross: A total of 9 genotypes yield red pigs, among which 4 are heterozygous for both genes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

PROBLEM 4 The following pedigree shows the inheritance of a rare, simple Mendelian dominant mutation with penetrance equal to 1/3. (A penetrance of 1/3 means that 1/3 of the individuals with the mutant genotype actually express the mutant phenotype.) The woman I-1 necessarily has the genotype Mm, where M is the mutant allele. Because the mutant allele is rare, you may assume that the male I-2 has the genotype mm. Individual II-1 is not affected. What is the probability that II-1 has the genotype Mm?

ANSWER This is a typical problem that makes use of Bayes' theorem. Let A be the event that individual II-1 has genotype Mm, and let B be the event that individual II-1 is not affected. Then event A′ is therefore the event that individual II-1 has the genotype mm. Because I-2 has genotype Mm, then Pr{A} = 1/2 and also Pr{A′} = 1/2. Now we can apply Bayes' theorem:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

where Pr{B|A} is the probability that an individual of genotype Mm is not affected and Pr{B|A′} is the probability that an individual of genotype mm is not affected. Because the penetrance is 1/3, then Pr{B|A} = 2/3. Because the genotype mm is never affected, P{B|A ′} = 1. Putting all this together, we obtain the answer we were seeking:

An alternative approach avoids the machinery of Bayes' theorem. Because individual I-1 has genotype Mm, the possible offspring of I-1 are of three types: (1) Mm and affected, with probability 1/2 × 1/3 = 1/6; (2) Mm and not affected, with probability 1/2 × 2/3 = 1/3; and (3) mm and not affected, with probability 1/2. Because we know that II-1 is not affected, possibility (1) can be ruled out. Thus, the probability that II-1 has genotype Mm, given that she is not affected, is given by (1/3)/[(1/3 + 1/2)] = 2/5, which agrees with the answer obtained with Bayes' theorem.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

ANALYSIS AND APPLICATIONS 3.1 In the cross Aa Bb Cc Dd × Aa Bb Cc Dd, in which all genes undergo independent assortment, what proportion of offspring are expected to be heterozygous for all four genes? 3.2 Consider a gene with four alleles, A1, A2, A3, and A4. In a cross between A1A2 and A3A4, what is the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

probability that a particular offspring inherits either A1 or A3 or both? 3.3 Assuming equal numbers of boys and girls, if a mating has already produced a girl, what is the probability that the next child will be a boy? If a mating has already produced two girls, what is the probability that the next child will be a boy? On which type of probability argument do you base your answers? 3.4 A cross is carried out between genotypes Aa BB Cc dd Ee and Aa Bb cc DD Ee. How many genotypes of progeny are possible?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

3.5 An individual of genotype AA Bb Cc DD Ee is testcrossed. Assuming that the loci undergo independent assortment, what fraction of the progeny are expected to have the genotype Aa Bb Cc Dd Ee? 3.6 Homozygous pea plants with round seeds are crossed to homozygous pea plants with wrinkled seeds. The F1 progeny undergo self-fertilization. A single round seed is chosen at random from the F2 progeny, and its DNA examined by electrophoresis as described in the text. What is the probability that the observed gel pattern will be B?

3.7 Assume that the trihybrid cross MM SS tt × mm ss TT is made in a plant species in which M and S are dominant but there is no dominance between T and t.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Consider the F2 progeny from this cross, and assume independent assortment. (a) How many phenotypic classes are expected? (b) What is the probability of genotype MM SS tt? (c) What proportion would be expected to be homozygous for all genes?

3.8 Shown here is a pedigree and a gel diagram indicating the clinical phenotypes with respect to phenylketonuria and the molecular phenotypes with respect to an RFLP that overlaps the PAH gene for phenylalanine hydroxylase. The individual II-1 is affected. (a) Indicate the expected molecular phenotype of II-1.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(b) Indicate the possible molecular phenotypes of II-2.

3.9 Which mode of inheritance is suggested by the following pedigree? Based on this hypothesis, and assuming that the trait is rare and has complete penetrance, what are the possible genotypes of all individuals in this pedigree?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.10 Huntington disease is a rare neurodegenerative human disease determined by a dominant allele, HD. The disorder is usually manifested after the age of 45. A young man has learned that his father has developed the disease. (a) What is the probability that the young man will later develop the disorder? (b) What is the probability that a child of the young man carries the HD allele?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

3.11 The accompanying gel diagram shows the banding patterns observed among the progeny of a cross between double heterozygous genotypes shown at the left (P1 and P2). On the right are shown the patterns of bands observed among the progeny. The bands are numbered 1–4.

(a) Identify which pairs of numbered bands correspond to the two segregating pairs of alleles. (b) Assuming that the two pairs of alleles undergo independent assortment, what is the probability that an

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

offspring of the cross shows the banding pattern in lane D?

3.12 Assume that the trait in the accompanying pedigree is due to simple Mendelian inheritance. (a) Is it likely to be due to a dominant allele or a recessive allele? Explain. (b) What is the meaning of the double horizontal line connecting III-1 with III-2? (c) What is the biological relationship between III-1 and III-2?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(d) If the allele responsible for the condition is rare, what are the most likely genotypes of all of the persons in the pedigree in generations I, II, and III? (Use A and a for the dominant and recessive alleles, respectively.)

3.13 In Drosophila, the mutant allele bwdts causing brown eyes (normal eyes are red) is temperature sensitive. In flies reared at 29°C the mutant allele is dominant, but in flies reared at 22°C the mutant allele is recessive. In a cross of bwdts/+ × bwdts/+, where the + sign denotes the wild-type allele of bwdts, what is the expected ratio of brown-eyed flies to red-eyed flies if the progeny are reared at 29°C? At 22°C?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

3.14 Pedigree analysis tells you that a particular parent may have the genotype AA BB or AA Bb, each with the same probability. Assuming independent assortment, what is the probability that this parent will produce an A b gamete? What is the probability that the parent will produce an A B gamete? 3.15 The pedigree shown here is for a rare autosomal recessive trait with complete penetrance. You may assume that no one in the pedigree has the recessive allele unless that person inherits it from either I-1 or II-4 or both.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(a) Ignoring the sibship in generation IV, what is the probability that both parents in the first-cousin mating are heterozygous? (b) Taking the sibship in generation IV into account, what is the probability that both parents in the first-cousin mating are heterozygous?

3.16 A clinical test for a certain disease is 100 percent accurate in individuals who are affected but also yields a false positive result in 0.2 percent of healthy individuals. If the proportion of affected individuals in a population is 0.002, what is the overall probability

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

that an individual selected at random will yield a positive test result? 3.17 The accompanying pedigree and gel diagram show the phenotypes of the parents for an SSRP that has multiple alleles. What are the possible phenotypes of the progeny, and in what proportions are they expected?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

3.18 The pollen from round pea seeds obtained from the F2 generation of a cross between a true-breeding strain with round seeds and a true-breeding strain with wrinkled seeds was collected and used en masse to fertilize plants from the true-breeding wrinkled strain. What fraction of the progeny is expected to have wrinkled seeds? (Assume equal fertility among all genotypes.) 3.19 Heterozygous Cp cp chickens express a condition called creeper, in which the leg and wing bones are shorter than normal (cp cp). The dominant Cp allele is lethal when homozygous. Two alleles of an independently segregating gene determine white (W −) versus yellow (ww) skin color. From matings between chickens heterozygous for both of these genes, which phenotypic classes will be represented

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

among the viable progeny, and what are their expected relative frequencies? 3.20 The F2 progeny from a particular cross exhibit a modified dihybrid ratio of 9 : 7 (instead of 9 : 3 : 3 : 1). What phenotypic ratio would be expected from a testcross of the F1 progeny? 3.21 In the mating Aa × Aa, what is the smallest number of offspring, n, for which the probability of at least one aa offspring exceeds 95 percent?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

3.22 A woman is affected with a trait due to a dominant mutant allele that shows 50 percent penetrance. If she has a child, what is the probability that it will be affected? 3.23 The pattern of coat coloration in dogs is determined by the alleles of a single gene, with S (solid) being dominant over s (spotted). Black coat color is determined by the dominant allele A of a second gene, and homozygous recessive aa animals are tan. A female having a solid tan coat is mated with a male having a solid black coat and produces a litter of six pups. The phenotypes of the pups are 2 solid tan, 2 solid black, 1 spotted tan, and 1 spotted black. What are the genotypes of the parents? 3.24 Consider a phenotype for which the allele N is dominant to the allele n. A mating Nn × Nn is carried out, and one individual with the dominant phenotype is chosen at random. This individual is testcrossed and the mating yields four offspring, each with the dominant phenotype. What is the probability that the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

parent with the dominant phenotype has the genotype Nn?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

3.25 Cystic fibrosis is a genetic disease resulting from homozygosity for one of a large number of mutant alleles. Consider the accompanying pedigree. Individual 2-2 is wishes to know whether she is a carrier (that is, is heterozygous) for a cystic fibrosis allele, so she undergoes genetic testing that will detect 90 percent of those alleles in a population. This test is negative. What is the probability that, despite these results, she is a carrier?

3.26 The accompanying gel diagram shows the phenotype of two parents (X and Y), each of whom is homozygous for two RFLPs that undergo independent assortment. Parent X has genotype A1A1 B1B1, where the A1 allele yields a band of 4 kb and the B1 allele yields a band of 6 kb. Parent Y has genotype A2A2 B2B2, where the A2 allele yields a band of 8 kb and the B2 allele yields a band of 2 kb. Show the expected phenotype of the F1 progeny as well as all possible phenotypes of the F2 progeny, along with their expected proportions.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

3.27 Complementation tests of the recessive mutant genes a through f produced the data in the accompanying matrix. The circles represent missing data. Assuming that all of the missing mutant combinations would yield data consistent with the entries that are known, complete the table by filling each circle with a + or − as needed.

3.28 In plants, certain mutant genes are known that affect the ability of gametes to participate in fertilization. Suppose that an allele A is such a mutation and that pollen cells bearing the A allele are only half as likely to survive and participate in fertilization as pollen cells bearing the a allele. Complete the Punnett square for the F2 generation in a monohybrid cross. What is the expected ratio of AA : Aa : aa plants in the F2 generation?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 Meiotic drive is an unusual phenomenon in which two alleles do not show Mendelian segregation from the heterozygous genotype. Examples are known from mammals, insects, fungi, and other organisms. The usual mechanism is one in which both types of gametes are formed, but one of them fails to function normally. The excess of the driving allele over the other can range from a small amount to nearly 100 percent. Suppose that D is an allele showing meiotic drive against its alternative allele d, and suppose that Dd heterozygotes produce functional D-bearing and d-bearing gametes in the proportions 3/4 : 1/4. In the mating Dd × Dd: (a) What are the expected proportions of DD, Dd, and dd genotypes? (b) If D is dominant, what are the expected proportions of D− and dd phenotypes? (c) Among the D− phenotypes, what is the ratio of DD : Dd?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(d) Answer parts (a) through (c), assuming that the meiotic drive takes place in only one sex. CHALLENGE PROBLEM 2 The accompanying table summarizes the effect of inherited tissue antigens on the acceptance or rejection of transplanted tissues, such as skin grafts, in mammals. The tissue antigens are determined in a codominant fashion, so that tissue taken from a donor of genotype Aa carries both the A and the a antigen. In the table, the + sign means that a graft of donor tissue is accepted by the recipient, and the − sign means that a graft of donor tissue is rejected by the recipient. The rule is:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Any graft will be rejected whenever the donor tissue contains an antigen not present in the recipient. In other words, any transplant will be accepted if, and only if, the donor tissue does not contain an antigen different from any already present in the recipient.

The diagram illustrated here shows all possible skin grafts between inbred (homozygous) strains of mice (P1 and P2) and their F1 and F2 progeny. Assume that the inbred lines P1 and P2 differ in only

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

one tissue-compatibility gene. For each of the arrows, what is the probability of acceptance of a graft in which the donor is an animal chosen at random from the population shown at the base of the arrow and the recipient is an animal chosen at random from the population indicated by the arrowhead?

CHALLENGE PROBLEM 3 A woman who carries a copy of the mutant allele of BRCA1 has a 60 percent chance of developing breast cancer during her lifetime, while a woman who is homozygous for the normal allele has a 12 percent risk (mutant homozygotes are inviable). In the accompanying pedigree, individual II-3 developed breast cancer and carries a mutant allele

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

of BRCA1. Her brother (II-2) is deceased, but married and had a daughter (III-1) with an unrelated woman who, based on genotype, carries two normal alleles. Based on the pedigree and the risk factors associated with the different genotypes, what is the probability that individual III-1 will develop breast cancer?

FOR FURTHER READING

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Mendel, Gregor Johann. (1866). Versuche über pflanzen-hybriden [Experiments concerning plant hybrids]. In Verhandlungen des naturforschenden Vereines in Brünn [Proceedings of the Natural History Society of Brünn], IV(1865), 3–47. Reprinted in Fundamenta Genetica, ed. Jaroslav Krˇízˇenecky', 15–56. Prague, Czech Republic: Czech Academy of Sciences, 1966. http://www.mendelweb.org/Mendel.html Mendel's original paper. Reid, J. B., & Ross, J. J. (2011). Mendel's genes: Toward a full molecular characterization. Genetics, 189(1), 3–10. http://doi.org/10.1534/genetics.111.132118

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

What we have learned about the molecular bases of the seven genes Mendel characterized. Sandell, M. A., & Breslin, P. A. S. (2006). Variability in a taste-receptor gene determines whether we taste toxins in food. Current Biology, 16(18), R792– R792. http://www.cell.com/currentbiology/pdf/S0960-9822(06)02061-6.pdf A short note describing the relationship between the PTC taster allele of hTASR38 and perception of bitterness in foods. Shoemaker, J. S., Painter, I. S., & Weir, B. S. (1999). Bayesian statistics in genetics: A guide for the uninitiated. Trends in Genetics, 15(9), 354–358. http://doi.org/10.1016/S0168-9525(99)01751-5

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

A concise introduction to the application of Bayesian probability to genetic problems.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:21.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

© Eric V. Grave/Science Source.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

CHAPTER 4 The Chromosomal Basis of Inheritance CHAPTER OUTLINE 4.1 The Stability of Chromosome Complements 4.2 Mitosis 4.3 Meiosis 4.4 Sex-Chromosome Inheritance 4.5 Probability in the Prediction of Progeny Distributions

4.6 Testing Goodness of Fit to a Genetic Hypothesis ROOTS OF DISCOVERY: Grasshopper, Grasshopper

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

E. Eleanor Carothers (1913) The Mendelian Ratio in Relation to Certain Orthopteran Chromosomes ROOTS OF DISCOVERY: The White-Eyed Male Thomas Hunt Morgan (1910) Sex-Limited Inheritance in Drosophila

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

LEARNING OBJECTIVES & SCIENCE COMPETENCIES Understanding how chromosomes and sex chromosomes are inherited as well as the basic principles of probability and statistical tests of hypotheses discussed in this chapter will allow you to satisfy the following science competencies: ■ Predict which products of mitosis or meiosis would result from normal chromosome behavior or chromosome misbehavior. ■ Recognize the characteristic pattern of X-linked inheritance, and be able to trace X chromosomes as they pass between the sexes from generation to generation.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Given a genetic cross, use the binomial distribution to calculate the probability of any particular combination of genotypes or phenotypes among the progeny. ■ Be able to formulate a genetic hypothesis, use it to predict expected results of a cross, and compare the expected results with observed results by means of a chi-square test for goodness of fit.

I

t came as no great revelation that genes are located in chromosomes. The parallel between their properties made this quite obvious:

1. Genes come in pairs; chromosomes come in pairs. 2. Alleles of a gene segregate; homologous chromosomes segregate.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

3. Unlinked genes undergo independent assortment; nonhomologous chromosomes undergo independent assortment. These parallels were recognized by Walter Sutton in 1902, based on his observations of spermatogenesis in grasshoppers (Brachystola magna); in his words, “the association of paternal and maternal chromosomes in pairs and their subsequent separation during the reducing division as indicated above may constitute the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

physical basis of the Mendelian law of heredity.” After that time, there was little doubt that chromosomes are the cellular carriers of the genes—but parallels do not constitute scientific proof, nor does widespread agreement among scientists. In this chapter, we examine some of the experimental evidence that was at the time— and still is—regarded as sufficient to prove the chromosome theory of heredity.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

INTERPHASE nucleus (left) of a cell in the bluebell (right) (genus Hyacinthoides). Source: (Left) © Pr. G. Giménez-Martín/Science Source, (Right) ©

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Selectphoto/Dreamstime.com.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.1 The Stability of Chromosome Complements The cell nucleus was first discovered in 1831, but not until the late 1860s was it understood that nuclear division nearly always accompanies cell division. The importance of the nucleus in inheritance was reinforced by the nearly simultaneous discovery

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

that the nuclei of two gametes fuse in the process of fertilization. The next major advance came in the 1880s with the discovery of chromosomes, which were easily visualized in the light microscope with the use of certain dyes. A few years later, chromosomes were found to segregate by an orderly process into the daughter cells formed by cell division as well as into the gametes formed by the division of reproductive cells. Three important regularities were observed about the chromosome complement (the complete set of chromosomes) of plants and animals: 1. The nucleus of each somatic cell (a cell of the body, in contrast to a germ cell, or gamete) contains a fixed number of chromosomes typical of the particular species. This number varies tremendously among species, and chromosome number bears little relation to the complexity of the organism (TABLE 4.1). 2. The chromosomes in the nuclei of somatic cells are usually present in pairs. For example, the 46 chromosomes of human beings consist of 23 pairs (FIGURE 4.1). Cells with nuclei that contain two similar sets of chromosomes are called diploid. A diploid individual carries two allelic copies of each gene present in each pair of chromosomes. The chromosomes occur in pairs because one chromosome of each pair derives from the maternal parent and the other from the paternal parent of the organism.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

3. The gametes that unite in fertilization to produce the diploid somatic cells have nuclei that contain only one set of chromosomes, consisting of one member of each pair. The gametic nuclei are haploid.

TABLE 4.1 Somatic (diploid) chromosome numbers of some plant and animal species Organism

Chromosome

Organism

number Field horsetail

216

Chromosome number

Yeast (Saccharomyces

32

cerevisiae) Bracken fern

116

Fruit fly (Drosophila

8

melanogaster) Giant sequoia

22

Nematode

11♂, 12♀

(Caenorhabditis

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

elegans) Macaroni wheat

28

House fly

12

Bread wheat

42

Scorpion

4

Fava bean

12

Geometrid moth

224

Garden pea

14

Common toad

22

Mustard cress

10

Chicken

78

Corn (Zea mays)

20

Mouse

40

Lily

24

Gibbon

44

Snapdragon

16

Human

46

(Arabidopsis thaliana)

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.1 Chromosome complement of a human male. There are 46 chromosomes, present in 23 pairs. At the stage of the division cycle in which these chromosomes were observed, each chromosome consists of two identical halves lying side by side longitudinally. Except for the members of one chromosome pair (the pair that determines sex), the members of all the chromosome pairs are the same color because they contain DNA molecules that were labeled with the same mixture of fluorescent dyes. The colors differ from one pair to the next because the dye mixtures differ in color. In some cases, the long and the short arms have been labeled with a different color. Courtesy of Michael R. Speicher, Institute of Human Genetics, Medical University of Graz.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

In multicellular organisms that develop from single cells, the presence of the diploid chromosome number in somatic cells and of the haploid chromosome number in germ cells indicates that there are two processes of nuclear division that differ in their outcome. One of these (mitosis) maintains the chromosome number; the other (meiosis) reduces the number by half. These two processes are examined in the following sections.

SUMMING UP ■ The behavior of chromosomes in gamete formation parallels that of alleles in Mendelian inheritance.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Somatic cells are typically diploid, whereas germ line cells are haploid.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.2 Mitosis Mitosis is a process of nuclear division that ensures that each of two daughter cells receives a diploid complement of chromosomes identical to the diploid complement of the parent cell. Mitosis is usually accompanied by cytokinesis, the process in which the cell itself divides to yield two daughter cells. The basic process of mitosis is remarkably uniform in all organisms:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1. Each chromosome is already present as a duplicated structure at the beginning of nuclear division. (The duplication of each chromosome coincides with the replication of the DNA molecule it contains.) 2. Each chromosome divides longitudinally into identical halves that separate from each other. 3. The separated chromosome halves move in opposite directions, and each becomes included in one of the two daughter nuclei that are formed. In a cell not undergoing mitosis, the chromosomes are invisible with a light microscope. Each consists of an elongated thread too thin to be seen. This stage of the cell cycle is called interphase. In preparation for mitosis, the DNA in the chromosomes is replicated during a period of interphase called S (FIGURE 4.2), which stands for synthesis of DNA. DNA replication is accompanied by chromosome duplication. Before and after S, there are periods, called G1 and G2, respectively, in which DNA replication does not take place. The cell cycle (the life cycle of a cell) is commonly described in terms of these three interphase periods followed by mitosis, M. The order of events is therefore G1 → S → G2 → M, as shown in Figure 4.2. In this representation, the M period also includes the division of the cytoplasm (cytokinesis) into two approximately equal parts, each containing one daughter nucleus.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

FIGURE 4.2 The cell cycle of a typical mammalian cell growing in tissue culture with a generation time of 24 hours.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The length of time required for a complete life cycle varies with cell type; it is 18–24 hours for the majority of cells in higher eukaryotes. The relative duration of the different periods in the cycle also varies considerably with cell type. Mitosis is usually the shortest period, requiring 0.5–2 hours. The cell cycle is an active, regulated process controlled by mechanisms that are essentially identical in all eukaryotes. These mechanisms are discussed in detail in the Molecular Genetics of the Cell Cycle and Cancer chapter. The transitions from G1 into S, and from G2 into M, are called checkpoints because they are delayed unless key processes have been completed (Figure 4.2). For example, at the G1/S checkpoint, either (in some cell types) sufficient time must have elapsed since the preceding mitosis or (in other cell types) the cell must have attained sufficient size for DNA replication to be initiated. Similarly, the G2/M checkpoint requires

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

that DNA replication and repair of any DNA damage be completed for the M phase to commence. Illustrated in FIGURE 4.3 are the essential features of chromosome behavior in mitosis. Mitosis is conventionally divided

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

into four stages: prophase, metaphase, anaphase, and telophase. It is important to stress that DNA replication (S phase) precedes meiosis, so that mitosis requires that those replicated chromosomes be correctly sorted into two daughter nuclei.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.3 Chromosome behavior during mitosis in an organism with two pairs of chromosomes (red/rose versus green/blue). At each stage, the smaller inner diagram represents the entire cell, and the larger diagram is an exploded view showing the chromosomes at that stage.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Prophase In interphase, the chromosomes have the form of extended filaments and cannot be seen with a light microscope as discrete bodies. Except for the presence of one or more conspicuous dark bodies (nucleoli), the nucleus has a diffuse, granular appearance.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The beginning of prophase is marked by the condensation of chromosomes to form visibly distinct threads within the nucleus. Chromatin condensation is brought about by large protein complexes known as condensins. At this stage, each chromosome is already longitudinally double, consisting of two closely associated subunits called chromatids. The longitudinally bipartite nature of each chromosome is readily seen later in prophase. Each pair of chromatids is the product of the duplication of one chromosome in the S period of interphase. The chromatids in a pair are held together at a specific region of the chromosome called the centromere. As prophase progresses, the chromosomes become shorter and thicker as a result of intricate coiling (see the Molecular Organization of Chromosomes and Genomes chapter). At the end of prophase, the nucleoli disappear and the nuclear envelope, a membrane surrounding the nucleus, abruptly disintegrates.

PROPHASE of Hyacinthoides © Pr. G. Giménez-Martín/Science Source

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Metaphase At the beginning of metaphase, the mitotic spindle forms. This spindle is an elongated, football-shaped array of spindle fibers consisting primarily of microtubules formed by polymerization of the protein tubulin. Many other proteins and at least one RNAprotein complex regulate tubulin polymerization and microtubule organization. The ends or poles of the spindle, where the microtubules converge, mark the locations of the centrosomes, which are the microtubule organizing centers where tubulin polymerization is initiated. Each pair of centrosomes results from the duplication of a single centrosome that takes place in interphase, followed by migration of the daughter centrosomes to opposite sides of the nuclear envelope.

METAPHASE of Hyacinthoides

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

© Pr. G. Giménez-Martín/Science Source

The spindle features three types of microtubules: (1) those that anchor the centrosome to the cell membrane, (2) those that arch between the centrosomes, and (3) those that become attached to the chromosomes. The manner in which these are established exemplifies an important organizing principle of biology known as exploration and stabilization. As microtubules are formed, new tubulin subunits are added to the growing end of the polymer, but the growing end can also become unstable and initiate a process of depolymerization in which the microtubule shrinks. Several proteins

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

help regulate the balance between polymerization and depolymerization. In spindle formation, microtubules grow out from the spindle poles in essentially random directions (the exploration part of the process), but unless something happens to stabilize the growing end, each polymer will ultimately undergo depolymerization and disappear. For the microtubules that become attached to the chromosomes, the event that stabilizes the growing end is contact with a structure known as the kinetochore, which coincides with the position of the centromere. The process of random exploration and stabilization thereby results in a situation in which only those chromosomal microtubules that make contact with a kinetochore become stabilized, while the others depolymerize. Analogous types of stabilization likely also account for the microtubules that attach to the cell membrane and those that arch between the centrosomes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

After the spindle fibers have become attached to the chromosomes, each chromosome moves to a position near the center of the cell where its kinetochore lies on an imaginary plane approximately equidistant from the spindle poles. This imaginary plane is called the metaphase plate. When aligned on the metaphase plate, the chromosomes reach their maximum condensation and are easiest to count and examine for differences in shape and appearance. Proper chromosome alignment is an important cell-cycle control checkpoint at metaphase in both mitosis and meiosis. In a cell in which a chromosome is attached to only one pole of the spindle, the completion of metaphase is delayed. The signal for proper chromosome alignment comes from the kinetochore, and the chemical nature of this signal seems to be the dephosphorylation of certain kinetochore-associated proteins. Through the signaling mechanism, when all of the kinetochores are under tension and

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

aligned on the metaphase plate, the metaphase checkpoint is passed and the cell continues the process of division.

Anaphase In anaphase, the proteins holding the chromatids together are dissolved. The centromeres become separate, and the two sister chromatids of each chromosome move toward opposite poles of the spindle. Once the centromeres separate, each sister chromatid

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

is regarded as a separate chromosome in its own right. Chromosome movement results in part from progressive shortening of the spindle fibers attached to the centromeres, which pulls the chromosomes in opposite directions toward the poles, and often also from a temporary elongation of the dividing cell in a direction paralleling the spindle. At the completion of anaphase, the chromosomes lie in two groups near opposite poles of the spindle. Each group contains the same number of chromosomes that was present in the original interphase nucleus.

ANAPHASE of Hyacinthoides. © Pr. G. Giménez-Martín/Science Source

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Telophase In telophase, a nuclear envelope forms around each compact group of chromosomes, nucleoli are formed, and the spindle disappears. The chromosomes undergo a reversal of condensation until they are no longer visible as discrete entities. The two daughter nuclei slowly assume a typical interphase appearance as the cytoplasm of the cell divides into two by means of a gradually deepening furrow around the periphery. (In plants, a new cell wall is synthesized between the daughter cells and separates them.)

TELOPHASE of Hyacinthoides © Pr. G. Giménez-Martín/Science Source

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

SUMMING UP ■ Mitosis, the process of somatic cell division, results in the production of two identical daughter nuclei. ■ In the cell cycle, DNA replication precedes mitosis. ■ Mitosis has four stages—prophase, metaphase, anaphase, and telophase.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.3 Meiosis Meiosis is the mode of cell division that results in haploid daughter cells containing only one member of each pair of chromosomes. Unlike mitosis, in which daughter cells are genetically identical, meiosis generates genetic diversity because each daughter cell contains a different set of alleles. Meiosis consists of two successive nuclear divisions. An overview of the chromosome

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

behavior is outlined in FIGURE 4.4.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.4 Overview of the behavior of a single pair of homologous chromosomes in meiosis. The key differences from

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

mitosis are the pairing of homologous chromosomes (A) and the two successive nuclear divisions (B and D) that reduce the chromosome number by half. These two stages are usually designated as meiosis I and meiosis II. For clarity, this diagram does not incorporate crossing over, an interchange of chromosome segments that takes place at the stage depicted in part A.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1. Prior to the first nuclear division, the members of each pair of homologous chromosomes become closely associated along their length (Figure 4.4). Each member of the pair is already replicated and consists of two sister chromatids joined at the centromere. The pairing of the homologous chromosomes therefore produces a four-stranded structure. 2. In the first nuclear division, the homologous chromosomes are separated from each other, with the members of each pair going to opposite poles of the spindle (Figure 4.4B). Each daughter chromosome consists of two chromatids attached to a common centromere (Figure 4.4C), so both of the two nuclei that are formed contain a haploid set of chromosomes. (Chromosomes are enumerated by counting the number of centromeres, not the number of chromatids.) 3. The second nuclear division loosely resembles a mitotic division, but there is no DNA replication. At metaphase, the chromosomes align on the metaphase plate; at anaphase, the chromatids of each chromosome are separated into opposite daughter nuclei (Figure 4.4D). The net effect of the two divisions in meiosis is the creation of four haploid daughter nuclei, each containing the equivalent of a single sister chromatid from each pair of homologous chromosomes (Figure 4.4E).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Figure 4.4 does not show that the paired homologous chromosomes can exchange genes. The exchanges result in the formation of chromosomes that consist of segments from one homologous chromosome intermixed with segments from the other. In Figure 4.4, the exchanged chromosomes would be depicted as segments of alternating color. The exchange process, which is one of the critical features of meiosis, is examined in the next section. In animals, meiosis takes place in specific cells called meiocytes—

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

a general term for the primary oocytes and spermatocytes in the gamete-forming tissues (FIGURE 4.5). The oocytes form egg cells, and the spermatocytes form sperm cells. Although the process of meiosis is similar in all sexually reproducing organisms, in the female of both animals and plants, only one of the four products develops into a functional cell while the other three disintegrate. In animals, the products of meiosis form either sperm or eggs.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

FIGURE 4.5 The life cycle of a typical animal. The number n is the number of chromosomes in the haploid chromosome complement. In males, the four products of meiosis develop into functional sperm; in females, only one of the four products develops into an egg.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In plants, the situation is slightly more complicated: 1. The products of meiosis typically form spores, which undergo one or more mitotic divisions to produce a haploid gametophyte organism. The gametophyte produces gametes by mitotic division of a haploid nucleus (FIGURE 4.6). 2. Fusion of haploid gametes creates a diploid zygote that develops into the sporophyte plant, which undergoes meiosis to produce spores and so restarts the cycle.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.6 The life cycle of corn, Zea mays. As is typical in higher plants, the diploid spore-producing (sporophyte) generation is conspicuous, whereas the gamete-producing (gametophyte) generation is microscopic. The egg-producing spore is the megaspore, and the sperm-producing spore is the microspore. Nuclei participating in meiosis and fertilization are shown in yellow and green, respectively.

In higher plants, such as corn and Mendel's peas, it is the sporophyte generation that accounts for most of the plant's material; the gametophyte is microscopic. In other plants, such as ferns, mosses, and liverworts, the situation is reversed, such that most of the plant consists of haploid gametophyte tissue. Meiosis is a more complex and considerably longer process than mitosis and usually requires days or even weeks to reach completion. FIGURE 4.7 provides an overview of the process. The essence of the process is that meiosis consists of two divisions of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

the nucleus but only one replication of the chromosomes. The nuclear divisions—called the first meiotic division and the second meiotic division—can be separated into a sequence of stages similar to those used to describe mitosis. The distinctive events of this important process, which occur during the first division of the nucleus, are described next.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.7 Chromosome behavior during meiosis in an organism with two pairs of homologous chromosomes (red/rose and green/blue). At each stage, the small diagram represents the entire cell and the larger diagram is an expanded view of the chromosomes at that stage.

The First Meiotic Division: Reduction The first meiotic division (meiosis I) is sometimes called the reductional division because it divides the chromosome number in half. By analogy with mitosis, the first meiotic division can be split into four stages, which are called prophase I, metaphase I, anaphase I, and telophase I. These stages are generally more

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

complex than their counterparts in mitosis. The stages and substages can be visualized with reference to Figure 4.7 and FIGURE 4.8.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.8 Substages of prophase I in microsporocytes of the lily (Lilium longiflorum). (A) Leptotene: Condensation of the chromosomes is initiated and beadlike chromomeres are visible along the length of the chromosomes. (B) Zygotene: Pairing (synapsis) of homologous chromosomes occurs (paired and unpaired regions can be seen particularly at the lower left in this photograph). (C) Early pachytene: Synapsis is completed and crossing over between homologous chromosomes occurs. (D) Late pachytene: Continuation of the shortening and thickening of the chromosomes. (E) Diplotene: Mutual repulsion of the paired homologous chromosomes, which remain held together at one or more cross points (chiasmata) along their length. Diakinesis follows (not shown): The chromosomes reach their maximum contraction. (F) Zygotene (at higher magnification in another cell) showing paired homologs and matching of chromomeres during synapsis. Parts A, B, C, E, and F courtesy of Marta Walters and Santa Barbara Botanic Garden, Santa Barbara, California. Part D courtesy of Herbert Stern. Used with permission of Ruth Stern.

Prophase I.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

This long stage lasts several days in most higher organisms. It is commonly divided into five substages: leptotene, zygotene, pachytene, diplotene, and diakinesis. These terms describe the appearance of the chromosomes at each substage. In the leptotene period, whose name literally means “thin thread,” the chromosomes first become visible as long, threadlike structures. The pairs of sister chromatids can be distinguished by electron microscopy. In this initial phase of chromosome

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

condensation, numerous dense granules appear at irregular intervals along their length. These localized contractions, called chromomeres, have a characteristic number, size, and position in a given chromosome (Figure 4.8A).

The zygotene period is marked by the lateral pairing, or synapsis, of homologous chromosomes, beginning at the chromosome tips. (The term zygotene means “paired threads.”) As the pairing proceeds in zipper-like fashion along the length of the chromosomes, it results in a precise chromomere-by-chromomere association (Figure 4.8B and F). Synapsis is facilitated by the synaptonemal complex, a protein structure that helps hold the aligned homologous chromosomes together. Each pair of synapsed homologous chromosomes is referred to as a bivalent.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Throughout the pachytene period (Figure 4.8C and D), whose name literally means “thick thread,” the chromosomes continue to shorten and thicken (Figure 4.7). By late pachytene, it can sometimes be seen that each bivalent (that is, each set of paired chromosomes) actually consists of a tetrad of four chromatids, but the two sister chromatids of each chromosome are usually juxtaposed very tightly. Genetic exchange by means of crossing over takes place during pachytene. In Figure 4.7, the sites of exchange are indicated by the points where chromatids of different

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

colors cross over each other.

At the onset of the diplotene period, the synaptonemal complex breaks down and the synapsed chromosomes begin to separate. Diplotene means “double thread,” and the diplotene chromosomes are now clearly double (Figure 4.8E). The pairs of homologous chromosomes remain held together at intervals by crossconnections resulting from crossing over. Each cross-connection is called a chiasma (plural, chiasmata) and is formed by a breakage and rejoining between nonsister chromatids. As shown in the chromosome and diagram in FIGURE 4.9, a chiasma results from physical exchange between chromatids of homologous chromosomes. In normal meiosis, each bivalent usually has at least one chiasma, and bivalents of long chromosomes often have three or more chiasmata.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

FIGURE 4.9 Light micrograph (A) and interpretive drawing (B) of a bivalent consisting of a pair of homologous chromosomes. This bivalent was photographed at late diplotene in a spermatocyte of a salamander, Oedipina poelzi. It shows two chiasmata where the chromatids of the homologous chromosomes appear to exchange pairing partners.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Courtesy of Dr. James Kezer. Used with permission by Dr. Stanley K. Sessions.

The final period of prophase I is diakinesis, in which the homologous chromosomes seem to repel each other and the segments not connected by chiasmata move apart. (Diakinesis means “moving apart.”) It is at this substage of prophase I that the chromosomes attain their maximum condensation. The homologous chromosomes in each bivalent remain connected by at least one chiasma, which persists until the first meiotic anaphase. Near the end of diakinesis, the formation of a spindle is initiated, and the nuclear envelope breaks down.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Metaphase I. Each bivalent is maneuvered into a position straddling the metaphase plate, with the centromeres of the homologous chromosomes oriented to opposite poles of the spindle (FIGURE 4.10A). The orientation of the centromeres determines which member of each bivalent will subsequently move to each pole, and whether the maternal or the paternal centromere is oriented toward

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

a particular pole is completely a matter of chance.

FIGURE 4.10 Later meiotic stages in microsporocytes of the lily Lilium longiflorum: (A) metaphase I; (B) anaphase I; (C) metaphase II; (D) anaphase II; (E) telophase II. Cell walls have begun to form in telophase, which will lead to the formation of four pollen grains. Courtesy of Herbert Stern. Used with permission of Ruth Stern.

As shown in FIGURE 4.11, the bivalents formed from nonhomologous pairs of chromosomes can be oriented on the metaphase plate in either of two ways. If each of the nonhomologous chromosomes is heterozygous for a pair of alleles,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

then one type of alignment results in A B and a b gametes, whereas the other type results in A b and a B gametes. Because the metaphase alignment takes place at random, the two types of alignment—and therefore the four types of gametes—are equally frequent. The ratio of the four types of gametes is 1 : 1 : 1 : 1, which means that the A, a and B, b pairs of alleles undergo independent assortment. That is, Genes on different chromosomes undergo

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

independent assortment because nonhomologous chromosomes align at random on the metaphase plate in meiosis I.

FIGURE 4.11 Independent assortment of genes (A or B) on nonhomologous chromosomes results from random alignment of nonhomologous chromosomes at metaphase I.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

ROOTS OF DISCOVERY Grasshopper, Grasshopper E. Eleanor Carothers (1913) University of Kansas, Lawrence, Kansas

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The Mendelian Ratio in Relation to Certain Orthopteran Chromosomes As an undergraduate researcher, Carothers showed that nonhomologous chromosomes undergo independent assortment in meiosis. For this purpose she studied a type of grasshopper in which one pair of homologous chromosomes had members of unequal length. At the first anaphase of meiosis in males, she could determine by observation whether the longer or the shorter chromosome went in the same direction as the X chromosome. As detailed in this paper, she found 154 of the former and 146 of the latter—a result in very close agreement with the 1 : 1 ratio expected from independent assortment. There is no mention of the Y chromosome because in the grasshopper she studied, the females have the sex chromosome constitution XX, whereas the males have the sex chromosome constitution X. In the males she examined, therefore, the X chromosome did not have a pairing partner.

“Another link is added to the already long chain of evidence Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

“Another link is added to the already long chain of evidence that the chromosomes are distinct morphological individuals continuous from generation to generation, and, as such, are the

”

bearers of the hereditary qualities.

The instrument referred to as a camera lucida was at that time in widespread use for studying chromosomes and other microscopic objects. It is an optical instrument containing a prism or an arrangement of mirrors that, when mounted on a microscope, reflects an image of the microscopic object onto a piece of paper where it can be traced.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The aim of this paper is to describe the behavior of an unequal bivalent in the primary spermatocytes of certain grasshoppers. The distribution of the chromosomes of this bivalent, in relation to the X chromosome, follows the laws of chance; and, therefore, affords direct cytological support of Mendel's laws. This distribution is easily traced on account of a very distinct difference in size of the homologous chromosomes. Thus, another link is added to the already long chain of evidence that the chromosomes are distinct morphological individuals continuous from generation to generation, and, as such, are the bearers of the hereditary qualities…. This work is based chiefly on Brachystola magna [a short-horned grasshopper]…. The entire complex of chromosomes can be separated into two groups, one containing six small chromosomes and the other seventeen larger ones. [One of the larger ones is the X chromosome.] Examination shows that this group of six small chromosomes is composed of five of about equal size and one decidedly larger. [One of the small ones is the homolog of the decidedly larger one, making this pair of chromosomes unequal in size.] … In early metaphases, the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

chromosomes appear as twelve separate individuals [the bivalents]. Side views show the X chromosome in its characteristic position near one pole…. Three hundred cells were drawn under the camera lucida to determine the distribution of the chromosomes in the asymmetrical bivalent in relation to the X chromosome…. In the 300 cells drawn, the smaller chromosome went to the same nucleus as the X chromosome 146 times, or in 48.7 percent of the cases; and the larger one, 154 times, or in 51.3 percent of the cases…. A consideration of the limited number of chromosomes and the large number of characters in any animal or plant will make it evident that each chromosome must control numerous different characters….

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Since the rediscovery of Mendel's laws, increased knowledge has been constantly bringing into line facts that at first seemed utterly incompatible with them. There is no cytological explanation of any other form of inheritance…. It seems to me probable that all inheritance is, in reality, Mendelian. Notice that Carothers did not make any association between X chromosome and sex—this understanding would come later. What was important was the contribution her work made to the growing body of evidence about the chromosomal basis of heredity. Source: E. E. Carothers, J. Morphol. 24 (1913): 487–511.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Anaphase I. In this stage, homologous chromosomes, each composed of two chromatids joined at an undivided centromere, separate from each other and move to opposite poles of the spindle (Figure 4.10B). Chromosome separation at anaphase is the cellular basis of the segregation of alleles:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The physical separation of homologous chromosomes in anaphase is the physical basis of Mendel's principle of segregation.

Note, however, that the centromeres of the sister chromatids are tightly stuck together and behave as a single unit. A specific protein acts as a glue holding the sister centromeres together. This protein appears in the centromeres and adjacent chromosome arms during S phase and persists throughout meiosis I. It disappears only at anaphase II, when sister-centromere cohesion is lost and the sister centromeres separate.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Telophase I. At the completion of anaphase I, a haploid set of chromosomes consisting of one homolog from each bivalent is located near each pole of the spindle (Figure 4.6). In telophase, the spindle breaks down, and, depending on the species, either a nuclear envelope briefly forms around each group of chromosomes or the chromosomes enter the second meiotic division after only a limited uncoiling.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The Second Meiotic Division: Equation The second meiotic division (meiosis II) is sometimes called the equational division because the chromosome number remains the same in each cell before and after the second division. In some species, the chromosomes pass directly from telophase I to prophase II without loss of condensation; in others, there is a brief pause between the two meiotic divisions and the chromosomes may “decondense” (uncoil) somewhat. Chromosome replication never takes place between the two divisions; the chromosomes present at the beginning of the second division are identical to those present at the end of the first. After a short prophase (prophase II) and the formation of seconddivision spindles, the centromeres of the chromosomes in each nucleus become aligned on the central plane of the spindle at metaphase II (Figure 4.10C). In anaphase II, the protein holding

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the sister centromeres together breaks down. As a result, the sister centromeres appear to split longitudinally, and the chromatids of each chromosome move to opposite poles of the spindle (Figure 4.10D). Once the centromeres split at anaphase II, each chromatid is considered to be a separate chromosome.

Telophase II (Figure 4.10E) is marked by a transition to the interphase condition of the chromosomes in the four haploid nuclei, accompanied by division of the cytoplasm. Thus, the second meiotic division superficially resembles a mitotic division. However, there is an important difference: The chromatids of a chromosome are usually not genetically identical along their entire length because of crossing over associated with the formation of chiasmata during prophase of the first division.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

SUMMING UP ■ Meiosis, the process of gamete formation, results in the production of four haploid nuclei. ■ In meiosis, there are two rounds of nuclear division: reductional (meiosis I) and equational (meiosis II). ■ Meiosis I results in the production of two haploid nuclei and the independent assortment of genes on different chromosomes. ■ Meiosis II parallels mitosis, in that centromeres divide, resulting in production of daughter haploid cells with identical genetic content.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ During prophase of meiosis I, homologous chromosomes form chiasmata, which results in the exchange of genetic material between them.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.4 Sex-Chromosome Inheritance The first rigorous experimental proof that genes are parts of chromosomes was obtained in experiments concerned with the pattern of transmission of the sex chromosomes, the chromosomes responsible for determination of the separate sexes in some plants and in nearly all animals. We will examine these results in this section.

Chromosomal Determination of Sex

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The sex chromosomes are an exception to the rule that all chromosomes of diploid organisms are present in pairs of morphologically similar homologs. As early as 1891, microscopic analysis had shown that one of the chromosomes in males of some insect species does not have a homolog. This unpaired chromosome was called the X chromosome, and it was present in all somatic cells of the males but in only half the sperm cells. The biological significance of these observations became clear when females of the same species were shown to have two X chromosomes. In other species in which the females have two X chromosomes, the male has one X chromosome along with a morphologically unmatched chromosome. The unmatched chromosome is referred to as the Y chromosome, and it pairs with the X chromosome during meiosis in males, usually along only part of its length because of a limited region of homology. The difference in chromosomal constitution between males and females is a chromosomal mechanism for determining sex at the time of fertilization. Whereas every egg cell contains an X chromosome, half of the sperm cells contain an X chromosome and the rest contain a Y chromosome. Fertilization of an X-bearing egg by an X-

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

bearing sperm results in an XX zygote, which normally develops into a female; fertilization by a Y-bearing sperm results in an XY zygote, which normally develops into a male (FIGURE 4.12). The result is a crisscross pattern of inheritance of the X chromosome in which a male receives his X chromosome from his mother and

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

transmits it only to his daughters.

FIGURE 4.12 The chromosomal basis of sex determination in mammals, many insects, and other animals.

The XX–XY type of chromosomal sex determination is found in mammals, including human beings, in many insects, and in other animals, as well as in some flowering plants. The female is called the homogametic sex because only one type of gamete (Xbearing) is produced. The male is called the heterogametic sex

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

because two different types of gametes (X-bearing and Y-bearing) are produced. When the union of gametes in fertilization is random, a sex ratio at fertilization of 1 : 1 is expected because males produce equal numbers of X-bearing and Y-bearing sperm. The X and Y chromosomes together constitute the sex chromosomes; this term distinguishes them from other pairs of chromosomes, which are called autosomes. Although the sex chromosomes control the developmental switch that determines the earliest stages of female or male development, the developmental process itself requires many genes scattered throughout the genome, including genes on the autosomes. The X chromosome also contains many genes with functions unrelated to sexual differentiation, as we will see in the next section. In most organisms, including human beings, the Y chromosome carries few genes other than those related to male determination.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

X-Linked Inheritance The compelling evidence that genes are located in chromosomes came from the study of a Drosophila gene for white eyes, which proved to be present in the X chromosome. Recall that in Mendel's crosses, it did not matter which trait was present in the male parent and which in the female parent: Reciprocal crosses gave the same result. One of the earliest exceptions to this rule was found by Thomas Hunt Morgan in 1910, in an early study of a mutant fruit fly that had white eyes (see Roots of Discovery: The White-Eyed Male). The wild-type eye color is a brick-red combination of red and brown pigments (FIGURE 4.13). Although white eyes can result from certain combinations of autosomal genes that eliminate the pigments individually, the white-eye mutation that Morgan studied results in a metabolic block that knocks out both pigments simultaneously. (The gene codes for a transmembrane protein that

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

is necessary to transport the eye pigment precursors into the pigment cells.)

FIGURE 4.13 Drawings (A) of a male and a female fruit fly, Drosophila melanogaster. The photographs (B) show the eyes of a wild-type red-eyed male and a mutant white-eyed male. (A) Illustrations © Carolina Biological Supply Company. Used with permission. Photographs courtesy of E. R. Lozovsky.

ROOTS OF DISCOVERY The White-Eyed Male Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Thomas Hunt Morgan (1910) Columbia University New York, New York

Sex-Limited Inheritance in Drosophila Morgan's genetic analysis of the white-eye mutation marks the beginning of Drosophila genetics. It is in the nature of science that as knowledge increases, the terms used to describe things change as well. This paper offers an example of this evolution, because the term sex-limited inheritance is used today to mean something completely different from Morgan's usage. What

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Morgan was referring to is now called X-linked inheritance or sex-linked inheritance. To avoid confusion, we have taken the liberty of substituting the modern equivalent wherever appropriate. Morgan was also unaware that Drosophila males had a Y chromosome. He thought that females were XX and males X, as in grasshoppers (see the Carothers paper). We have also supplied the missing Y chromosome. Morgan's gene symbols have been retained as in the original. He uses R for the wild-type allele for red eyes and W for the recessive allele for white eyes. This is a curious departure from the convention, already introduced by Mendel, that dominant and recessive alleles should be represented by the same symbol. Today, we use w for the recessive allele and w1 for the dominant allele.

“No white-eyed females appeared.” The experiments Morgan conducted were based on the spontaneous appearance of a single white-eyed male in an otherwise true-breeding wild-type (red-eyed) strain. When that male was mated with its wild-type sisters, all of the progeny had red eyes. When these F1 progeny were crossed:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

[T]he F1 hybrids, inbred, produced

No white-eyed females appeared. The new character showed itself to be sex-linked in the sense that it was transmitted only to the grandsons. But was the white-eyed phenotype sex-limited in the modern sense, restricted in its expression, occurring only in males?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

That hypothesis was ruled out by Morgan's subsequent discovery: The white-eyed male (mutant) was later crossed with some of his daughters (F1), and produced

The results show that the new character, white eyes, can be carried over to the females by a suitable cross, and is in consequence in this sense not limited to one sex. It will be noted that the four classes of individuals occur in approximately equal numbers (25 percent)….

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Based on these results, Morgan hypothesized the existence of “factors” that determine sex (X and what we now know as Y) and eye color (R for red and W for white). In addition, he determined that: [T]he spermatozoa of the white-eyed male carry the “factor” for white eyes “W”; … half of the spermatozoa carry a sex factor “X,” [and] the other half lack it, i.e., the male is heterozygous for sex. [The male is actually XY.] Thus, the symbol for the male is “WXY,” and for his two kinds of spermatozoa WX—Y. Assume that all of the eggs of the red-eyed female carry the red-eyed “factor” R; and that all of the eggs (after meiosis) carry one X each, the symbol for the red-eyed female will be, therefore, RRXX and that for her eggs will be RX…. Morgan then described four additional crosses (mostly backcrosses), in which the results obtained matched those predicted by the hypothesis. These findings led him to the critical conclusion:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

In order to obtain these results it is necessary to assume that, when the two classes of spermatozoa are formed in the RXY male, R and X go together…. The fact is that this R and X are combined and have never existed apart. Interestingly, in the original paper, Morgan never uses the word “chromosome.” Genetic elements are referred to as “factors,” and this is reflected in the nomenclature he used. In modern terminology, the X chromosomes carrying the alternative alleles would be designated as Xw+ (red) and Xw (white). Taken in combination with Carothers's cytological observations, Morgan's hypothesis suggests that his X “factor” is indeed the X chromosome as we know it today. Source: T. H. Morgan, Science 32 (1910): 120–122.

Morgan's study started with a single male with white eyes that appeared in a wild-type laboratory population that had been maintained for many generations. In a mating of this male with wildtype females, all of the F1 progeny of both sexes had red eyes, which showed that the allele for white eyes is recessive. In the F2

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

progeny from the mating of F1 males with females, Morgan observed 2459 red-eyed females, 1011 red-eyed males, and 782 white-eyed males. It was clear that the white-eyed phenotype was somehow connected with sex, because all of the white-eyed flies were males. White eyes, however, were not restricted to males. For example, when red-eyed F1 females from the cross of wild-type ♀ × white ♂ were backcrossed with their white-eyed fathers, the progeny consisted of both red-eyed and white-eyed females and red-eyed and white-eyed males in approximately equal numbers.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

A key observation came from the mating of white-eyed females with wild-type males. All of the female progeny had wild-type eyes, but all of the male progeny had white eyes. This is the reciprocal of the original cross of wild-type ♀ × white ♂, which had given only wild-type females and wild-type males, so the reciprocal crosses gave different results. Morgan realized that reciprocal crosses would yield different results if the allele for white eyes was present in the X chromosome. A gene on the X chromosome is said to be X-linked. The normal chromosome complement of Drosophila melanogaster is shown in FIGURE 4.14. Females have an XX chromosome complement, whereas males are XY. Morgan's hypothesis was that an X chromosome contains either a wild-type w+ allele or a mutant w allele, whereas the Y chromosome does not contain a counterpart of the white gene. Using the white allele present in an X chromosome to represent the entire X chromosome, we can write the genotype of a white-eyed male as wY and that of a wildtype male as w+Y. Because the w allele is recessive, white-eyed females are of genotype ww and wild-type females are either heterozygous w+w or homozygous w+w+. The implications of this

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

model for reciprocal crosses are shown in FIGURE 4.15. The mating wild-type ♀ × white ♂ is cross A, and that of white ♀ × wild-type ♂ is cross B.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.14 The diploid chromosome complements of a male and a female Drosophila melanogaster. The centromere of the X chromosome is nearly terminal, but that of the Y chromosome divides the chromosome into two unequal arms. The large autosomes (chromosomes 2 and 3, shown in blue and green) are not easily distinguishable in these types of cells. The tiny autosome (chromosome 4, shown in yellow) appears as a dot.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

FIGURE 4.15 Chromosomal interpretation of the results obtained in F1 and F2 progenies in crosses of Drosophila. Cross A is a mating of a wild-type (red-eyed) female with a white-eyed male. Cross B is the reciprocal mating of a white-eyed female with a red-eyed male. In the X chromosome, the wild-type w+ allele is shown in red,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the mutant w allele in white. The Y chromosome does not carry either allele of the w gene.

The X-linked mode of inheritance does account for the different phenotypic ratios observed in the F1 and F2 progeny from the crosses. The characteristics of X-linked inheritance can be summarized as follows: 1. Reciprocal crosses resulting in different phenotypic ratios in the sexes often indicate X-linked inheritance. In the case of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

white eyes in Drosophila, the cross of a red-eyed female with a white-eyed male yields all red-eyed progeny (Figure 4.15, cross A), whereas the cross of a white-eyed female with a red-eyed male yields red-eyed female progeny and whiteeyed male progeny (Figure 4.15, cross B). 2. Heterozygous females transmit each X-linked allele to approximately half of their daughters and half of their sons; this is illustrated in the F2 generation of cross B in Figure 4.15. 3. Males that inherit an X-linked recessive allele exhibit the recessive trait because the Y chromosome does not contain a wild-type counterpart of the gene. Affected males transmit the recessive allele to all of their daughters but none of their sons; this principle is illustrated in the F1 generation of cross A in Figure 4.16. Any male that is not affected must carry the wild-type allele in his X chromosome.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The essence of X-linked inheritance is captured in the Punnett square in FIGURE 4.16: A male transmits his X chromosome only to his daughters, whereas a female transmits an X chromosome to offspring of both sexes.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

FIGURE 4.16 Punnett square showing that, in chromosomal sex determination, each son gets his X chromosome from his mother and his Y chromosome from his father.

Pedigree Characteristics of Human XLinked Inheritance An example of a human trait with an X-linked pattern of inheritance in humans is hemophilia A, a severe disorder of blood clotting determined by a recessive allele. Affected persons lack a bloodclotting protein called factor VIII that is needed for normal clotting, and they suffer excessive, often life-threatening bleeding after

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

injury. A famous pedigree of hemophilia starts with Queen Victoria of England (FIGURE 4.17). One of her sons, Leopold, was hemophilic, and two of her daughters were heterozygous carriers of the gene. Two of Victoria's granddaughters were also carriers, and by marriage they introduced the gene into the royal families of Russia and Spain. The heir to the Russian throne of the Romanovs, Tsarevich Alexis, was afflicted with the condition. He inherited the gene from his mother, the Tsarina Alexandra, one of Victoria's granddaughters. The Tsar, the Tsarina, Alexis, and his four sisters

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

were all executed by the Russian Bolsheviks in the 1918 revolution. The present royal family of England is descended from a normal son of Victoria and is free of the disease.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.17 Genetic transmission of hemophilia A among the descendants of Queen Victoria of England, including her granddaughter, Tsarina Alexandra of Russia, and Alexandra's five children. The photograph is that of Tsar Nicholas II, Tsarina Alexandra, and their children. Tsarevich Alexis was afflicted with hemophilia. (inset) Photo courtesy of the Trustees of the Boston Public Library, Print Department.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

X-linked inheritance in human pedigrees has several characteristics that distinguish it from other modes of genetic transmission: ■ For any rare trait due to an X-linked recessive allele, the affected individuals are exclusively, or almost exclusively, male. There is an excess of males because females carrying the rare X-linked recessive are almost exclusively heterozygous and so do not express the mutant phenotype. ■ Affected males who reproduce have normal sons. This follows from the fact that a male transmits his X chromosome only to his daughters. ■ A woman whose father was affected has normal sons and affected sons in the ratio 1 : 1. This is true because any daughter of an affected male must be heterozygous for the recessive allele.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Heterogametic Females In some organisms, the homogametic and heterogametic sexes are reversed; that is, the males are XX and the females are XY. This type of sex determination is found in birds, in some reptiles and fish, and in moths and butterflies. The reversal of XX and XY in the sexes results in an opposite pattern of nonreciprocal inheritance of X-linked genes. To distinguish sex determination in these organisms from the usual XX–XY mechanism, the sex chromosome constitution in the homogametic sex is designated ZZ and that in the heterogametic sex as WZ. Hence, in organisms with heterogametic females, the chromosomal constitution of the females is designated WZ and that of the male ZZ. A specific example of Z-linked inheritance in chickens is shown in FIGURE 4.18. A few breeds of chickens have feathers with alternating transverse bands of light and dark color, resulting in a

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

phenotype known as barred. In other breeds, the feathers are uniformly colored and nonbarred. Reciprocal crosses between truebreeding barred and true-breeding nonbarred breeds give the results shown in Figure 4.18, which indicate that the gene that determines barring must be dominant and must be located in the Z

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

chromosome.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.18 Barred feathers in chickens—a classic example showing that chromosomal sex determination in birds is the reverse of that in mammals. In birds, females are the heterogametic sex. The Z chromosome carrying the dominant barred mutation is indicated in red.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Nondisjunction as Proof of the Chromosome Theory of Heredity The parallel between the inheritance of the Drosophila white mutation and the genetic transmission of the X chromosome supported the chromosome theory of heredity that genes are parts of chromosomes. Other experiments with Drosophila provided the definitive proof.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

One of Morgan's students, Calvin Bridges, discovered rare exceptions to the expected pattern of inheritance in crosses with several X-linked genes. For example, when white-eyed Drosophila females were mated with red-eyed males, most of the progeny consisted of the expected red-eyed females and white-eyed males. However, about 1 in every 2000 F1 flies was an exception—that is, either a white-eyed female or a red-eyed male. Bridges showed that these rare exceptional offspring resulted from occasional failure of the two X chromosomes in the mother to separate from each other during meiosis—a phenomenon called nondisjunction. The consequence of nondisjunction of the X chromosome is the formation of some eggs with two X chromosomes and others with none. Four classes of zygotes are expected from the fertilization of these abnormal eggs (FIGURE 4.19). Animals with no X chromosome are not detected because embryos that lack an X are not viable; likewise, most progeny with three X chromosomes die early in development. Microscopic examination of the chromosomes of the exceptional progeny from the cross white♀ × wild-type ♂ showed that the exceptional white-eyed females had two X chromosomes plus a Y chromosome, and the exceptional red-eyed males had a single X but were lacking a Y. The latter, with a sex-chromosome constitution denoted XO, were sterile males.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.19 Results of nondisjunction of the X chromosomes in the first meiotic division in a female Drosophila.

These and related experiments demonstrated conclusively the validity of the chromosome theory of heredity. Chromosome theory of heredity: Genes are physically located within chromosomes. Bridges's evidence for the chromosome theory was that exceptional behavior on the part of chromosomes is precisely

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

paralleled by exceptional inheritance of their genes. This proof of the chromosome theory ranks among the most important and elegant experiments in genetics.

Sex Determination in Drosophila In the XX–XY mechanism of sex determination, the Y chromosome is associated with the male. In some organisms, including humans, this association occurs because the presence of the Y

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

chromosome triggers events in embryonic development that result in the male sexual characteristics (see the Human Karyotypes and Chromosome Behavior chapter). Drosophila is unusual among organisms with an XX–XY type of sex determination in that the Y chromosome, although associated with maleness, is not male determining. We already saw (Figure 4.19) that individuals with the genotype XXY are female; in fact, they are fully viable and fertile. TABLE 4.2 shows the relationship between sex-chromosome genotype, number of sets of autosomes, and phenotypic sex. XXY embryos in Drosophila develop into morphologically normal, fertile females, whereas XO embryos develop into morphologically normal, but sterile males. (The O is written in the formula XO to emphasize that a sex chromosome is missing.) The sterility of XO males shows that the Y chromosome, though not necessary for male development, is essential for male fertility; in fact, the Drosophila Y chromosome contains six genes required for the formation of normal sperm.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

TABLE 4.2 Relationship of sex-chromosome genotype, autosome sets, and phenotypic

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

sex in Drosophila Sex chromosomes

Sets of autosomes

X:A ratio

Phenotypic sex

XX

2

1

Female

XY

2

0.5

Male

XXY

2

1

Female

XO

2

0.5

Male (sterile)

XX

3

0.67

Intersex

The genetic determination of sex in Drosophila depends on an Xlinked gene known as sex-lethal (Sxl) because some mutant alleles are lethal when present in males. As shown in FIGURE 4.20, the Sxl gene is active in normal females and inactive in normal males. The product of Sxl is the sex-lethal protein SXL—an RNA-binding protein that binds with the RNA transcripts of several genes and causes female-specific RNA processing. In the absence of SXL, these transcripts undergo male-specific processing. Hence the targets of SXL are genes whose protein products are normally made only in females. One of the targets of SXL is the Sxl gene itself, so a small burst of SXL activity early in female development leads to self-sustaining production of SXL because of feedback in the processing of more Sxl transcripts.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

FIGURE 4.20 Early steps in the genetic control of sex determination in Drosophila through the activity of the sex-lethal gene Sxl.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

What makes for the early burst of SXL is the amount of product produced by four X-linked genes known as X-linked signal elements (XSE) (Figure 4.20). If the amount of XSE is high, the early burst occurs and the organism becomes female. In contrast, if the amount of XSE is low, the early burst does not occur and the organism becomes male (Figure 4.20). For almost 100 years, geneticists believed that sex was determined by the ratio of the number of X chromosomes (X) relative to the number of sets of autosomes (A). As we see in Table 4.2, the correlation is perfect. In normal females (XXAA), the ratio equals 1; in normal males (XAA), it equals 1/2; and in rare individuals of composition XXAAA, the ratio is 2/3 and the organism develops as an intersex. Furthermore, rare XA haploids initially develop as female, although they eventually die, which again supports the idea that an embryo with a ratio of X : A of 1/2 develops as male, an embryo with an X : A ratio of 1 develops as a female, and an embryo with an X : A ratio between 1/2 and 1 develops as an intersex.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

While the correlation with X : A is perfect, recent data show that this hypothesis is actually wrong. Sex in Drosophila is based on Sxl reacting to the amount of XSE gene products, which in normal flies is determined by the number of X chromosomes. The autosomes do play a role, but not in sex determination. Their role focuses on determining when in early development the commitment to sexual differentiation takes place. In normal XXAA or XAA embryos with two sets of autosomes, commitment takes place after 15 cycles of division. In XA haploid embryos with one set of autosomes, commitment takes place one cell cycle later, when the amount of XSE is sufficient to induce Sxl. In XXAAA embryos, some cells commit in cycle 13 and have too little XSE to induce Sxe, whereas other cells commit in cycle 14 and have enough XSE to induce Sxl. The result is an embryo that is a mosaic of male and female cells, and the phenotypic result is an intersex.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Sex determination in Drosophila is an excellent example of why correlation does not necessarily mean causation. It also illustrates the principle that ideas in science—even those that are long entrenched and widely believed—may sometimes turn out to be wrong.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

SUMMING UP ■ Sex chromosomes are dimorphic. ■ In species with chromosomal sex determination, one sex is homogametic (XX females in humans) and the other is heterogametic (XY males in humans). The opposite hold in some other organisms (including birds): Males are homogametic and females are heterogametic. ■ Genes on the X chromosome display a “crisscross” pattern of inheritance. ■ In pedigrees, X-linked recessive traits occur primarily in male offspring of unaffected parents. ■ The occurrence of nondisjunction of X chromosomes in Drosophila led to the proof that genes lie on chromosomes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ In Drosophila, sex is determined by a cascade of events that is triggered by the number of X chromosomes in the developing embryo.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.5 Probability in the Prediction of Progeny Distributions Genetic transmission includes a large component of chance. A particular gamete from an Aa organism might or might not include the A allele, depending on chance. A particular gamete from an Aa Bb organism might or might not include both the A and B alleles, depending on the chance orientation of the chromosomes on the metaphase I plate. Genetic ratios result not only from the chance assortment of genes into gametes, but also from the chance combination of gametes into zygotes. Although exact predictions are not possible for any particular event, it is possible to determine the probability that a particular event will be realized, as we saw in the Mendelian Genetics: The Principles of Segregation and Assortment chapter. In this section, we consider some additional probability methods used in interpreting genetic data.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Using the Binomial Distribution in Genetics The addition rule of probability deals with outcomes of a genetic cross that are mutually exclusive. Outcomes are “mutually exclusive” if they are incompatible in the sense that they cannot occur at the same time. For example, the possible sex distributions among three children consist of four mutually exclusive outcomes: zero, one, two, or three girls. These outcomes have probabilities 1/8, 3/8, 3/8, and 1/8, respectively. The addition rule states that the overall probability of any combination of mutually exclusive events is equal to the sum of the probabilities of the events taken separately. For example, the probability that a sibship of size three contains at least one girl includes the outcomes one, two, and three girls, so the overall probability of at least one girl equals 3/8 + 3/8 + 1/8 = 7/8.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

The multiplication rule of probability deals with outcomes of a genetic cross that are independent. Any two possible outcomes are independent if the knowledge that one outcome is actually realized provides no information about whether the other is also realized. For example, in a sequence of births, the sex of any particular child is not affected by the sex of any sibling born earlier; moreover, it has no influence whatsoever on the distribution of sexes of any siblings born later. Each successive birth is independent of all the others. When possible outcomes are independent, the multiplication

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

rule states that the probability of any combination of outcomes being realized equals the product of the probabilities of all the individual outcomes taken separately. For example, the probability that a sibship of three children will consist of three girls equals 1/2 × 1/2 × 1/2, because the probability of each birth resulting in a girl is 1/2, and the successive births are independent. Probability calculations in genetics frequently use the addition and multiplication rules together. For example, the probability that all three children in a family will be of the same sex uses both the addition and multiplication rules. The probability that all three will be girls is (1/2)(1/2)(1/2) = 1/8, and the probability that all three will be boys is also 1/8. Because these outcomes are mutually exclusive (a sibship of size three cannot include three boys and three girls), the probability of either three girls or three boys is the sum of the two probabilities, or 1/8 + 1/8 = 1/4. The other possible outcomes for sibships of size three are that two of the children will be girls and the other a boy, and that two will be boys and the other a girl. For each of these outcomes, three different orders of birth are possible—for example, GGB, GBG, and BGG—each having a probability of 1/2 × 1/2 × 1/2 = 1/8. The probability of two girls and a boy, disregarding birth order, is the sum of the probabilities for the three possible orders, or 3/8; likewise, the probability of two

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

boys and a girl is also 3/8. Therefore, the distribution of probabilities for the sex ratio in families with three children is:

The sex ratio information in this display can be obtained more directly by expanding the binomial expression (p + q)n, in which p is the probability of the birth of a girl (1/2), q is the probability of the birth of a boy (1/2), and n is the number of children. In the present example,

in which the red numerals are the possible number of birth orders for each sex distribution. Similarly, the binomial distribution of probabilities for the sex ratios in families of five children is:

Each term tells us the probability of a particular combination. For example, the third term is the probability of three girls (p3) and two

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

boys (q2) in a family that has five children:

There are n + 1 terms in a binomial expansion to the power n. The exponents of p decrease by one from n in the first term to 0 in the last term, and the exponents of q increase by one from 0 in the first term to n in the last term. The coefficients generated by successive values of n can be arranged in a regular triangle known as Pascal's triangle, shown for n = 0 to 10 in FIGURE 4.21. The horizontal rows of the triangle are symmetrical; each row begins

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

and ends with a 1, and each other entry can be obtained as the sum of the two numbers on either side of it in the row above.

FIGURE 4.21 Pascal's triangle. The numbers in the nth row are the coefficients of the terms in the expansion of the polynomial (p + q)n.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

To generalize just a bit, if the probability of event A is p, and that of event B is q, and if the two events are independent and mutually exclusive, then the probability that A will be realized four times and B two times (in a specified order) is, by the multiplication rule, p4q2. Usually we are interested in a combination of events regardless of their order, such as “four A and two B.” In this case, we multiply the probability that the combination 4A : 2B will be realized in any specified order by the number of possible orders. The number of different combinations of six events, four of type A and two of type B, is given by the coefficient of p4q2 in the expansion of (pA + qB)6. This coefficient can be found in the row for n = 6 in Pascal's triangle, as the fifth entry from the left. (It is the fifth because the successive entries are the coefficients of p0q6, p1q5, p2q4, p3q3, p4q2, p5q1, and p6q0.) Hence, the overall probability of four

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

realizations of A and two realizations of B in six trials is given by 15p4q2. The general rule for repeated trials of events with constant probabilities is as follows: If the probability of event A is p and the probability of the alternative event B is q, the probability that, in n trials, event A is realized s times and event B is realized t times is

in which s + t = n and p + q = 1.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The symbol n! is read as “n factorial,” and it stands for the product of all positive integers from 1 through n (that is, n! = 1 × 2 × 3 × … × n). Values n factorial from n = 0 through n = 15 are given in TABLE 4.3. The value 0! = 1 is defined arbitrarily to generalize its use in mathematical formulas. The magnitude of n! increases very rapidly; 15! is more than 1 trillion.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

TABLE 4.3 Factorials n

n!

n

n!

0

1

8

40,320

1

1

9

362,880

2

2

10

3,628,800

3

6

11

39,916,800

4

24

12

479,001,600

5

120

13

6,227,020,800

6

720

14

87,178,291,200

7

5040

15

1,307,674,368,000

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Equation 1 applies even when either s or t equals 0, because 0! = 1. (Remember also that any number raised to the zero power equals 1.) Any individual term in the expansion of the binomial (p + q)n is given by Equation 1) for the appropriate values of s and t. In Pascal's triangle, successive entries in the nth row are the values of n!/(s! t!) for s = 0, 1, 2, …, n. Let us consider a specific application of Equation 1, in which we calculate the probability that the four kittens shown in the photo are the result of a mating between two heterozygous parents that yielded exactly the expected 3 : 1 ratio of the dominant (gray) and recessive (white) traits among sibships of a particular size. The probability p of a kitten showing the dominant trait is 3/4, and the probability q of a kitten showing the recessive trait is 1/4. This is the “expected” Mendelian ratio. In this case, n = 4, s = 3, t = 1, and the probability of this combination of events is:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

That is, in only 42 percent of the cat families with four kittens would the offspring exhibit the expected 3 : 1 phenotypic ratio; the other sibships would deviate in one direction or the other because of chance variation. The importance of this example lies in its demonstration that although a 3 : 1 ratio is the “expected” outcome (and the single most likely outcome), the majority of the families (58 percent) actually have a distribution of offspring different from 3 : 1.

© Iv Mirin/Shutterstock.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Meaning of the Binomial Coefficient The factorial part of the binomial expansion in Equation 1, which equals n!/(s! t!), is called the binomial coefficient. As we have noted, this ratio enumerates all possible ways in which s elements of one kind and t elements of another kind can be arranged in order, provided that the s elements and the t elements are not distinguished among themselves. A specific example might include s yellow peas and t green peas. Although the yellow peas and the green peas can be distinguished from each other because they have different colors, the yellow peas are not distinguishable from one another (because they are all yellow); the same constraint applies to the green peas (because they are all green).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

The reasoning behind the factorial formula begins with the observation that the total number of elements is s + t = n. Given n elements, each distinct from the next, the number of different ways in which they can be arranged is:

Why? Because the first element can be chosen in n ways, and

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

once it is chosen, the next can be chosen in n − 1 ways (because only n − 1 are left to choose from). Once the first two are chosen, the third can be chosen in n − 2 ways, and so on. Finally, once n − 1 elements have been chosen, there is only one way to choose the last element. The s + t elements can be arranged in n! ways, provided that the elements are all distinguished among themselves. However, applying again the argument we just used, each of the n! particular arrangements must include s! different arrangements of the s elements and t! different arrangements of the t elements, or s! × t! altogether. Dividing n! by s! × t! therefore yields the binomial coefficient for the number of ways in which the s elements and the t elements can be arranged when the elements of each type are not distinguished among themselves.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

SUMMING UP ■ The binomial distribution forms the basis for making predictions based on genetic hypotheses. ■ The probability of independent events occurring is the product of the probability of each event occurring by itself. ■ The probability of mutually exclusive events occurring simultaneously is the sum of the events occurring by themselves.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The binomial equation can be used to predict the exact probability of a particular outcome of a genetic cross occurring.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.6 Testing Goodness of Fit to a Genetic Hypothesis Geneticists often need to decide whether an observed ratio is in satisfactory agreement with a theoretical prediction. Mere inspection of the data is unsatisfactory because different investigators may disagree. What we need is to be able to set some threshold—that is, some objective standard for determining whether observed results agree with our hypothesis. To understand how this is done, we need to investigate the implications of probability theory a bit further.

Random Variables and Distributions Before we tackle testing genetic hypotheses, let's consider the simplest possible example of a binomial variable—a coin flip. Suppose we were to flip a coin 20 times, and the result was 18 heads. Given the hypothesis that it is a fair coin, our expected number of heads, of course, is 10; the question is whether our observed outcome differs so much from that expectation that we would reject the hypothesis that it is a fair coin.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

One approach to this question would be to apply equation 1 as we did for the four kittens. For this example, we know that

So the probability of observing this outcome is

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

In simple terms, the probability of getting this exact outcome is rather small. But let's frame the question a bit differently. Suppose we wanted to know, if we repeated the experiment many times (say 1000), what the distribution of outcomes look like. Of course, no one wants to take the time to do 20 coin flips that many times, so as an alternative, we'll ask a computer to simulate this scenario for us. The outcome is shown in FIGURE 4.22 as a histogram of the outcomes of the 1000 simulated experiments. Notice the lower and upper parts of the distribution, colored in red, where the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

number of heads observed is either 6 or less or 14 or greater. In fact, these outcomes, taken together, account for the 5 percent of the distribution that differs the most from the expected outcome (in this case, 10 heads).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 4.22 Histogram showing the outcomes of 1000 simulations of flipping a coin 20 times. Outcomes shown in red represent the 5 percent of the total that differ most from the expected number of 10.

Why is this important? Let's return to our original experiment and frame it as an assessment of two mutually exclusive hypotheses: H0: The coin is fair (Pr{Head} = 0.5}. H1: The coin is not fair (Pr{Head ≠ 0.5}. H0 is called the null hypothesis and H1 is called the alternative hypothesis. Statisticians would then ask whether the deviation of observed results from those expected under the null hypothesis are too great to be explained by chance. What is “too great”? In this

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

case (and other cases commonly used in genetics), the cut-off is defined as deviation from the null hypothesis that would be expected to occur by chance less than 5 percent of the time. We refer to that deviation as the critical value. In the case of the experiment depicted in Figure 4.22, that would be any of the outcomes that fall in the values shown in red (less than 7 or more than 13 heads). Given that our original outcome (18 heads) falls into that category, we would reject the null hypothesis. A test of this nature is called a goodness of fit test, where the word “fit” means how closely the observed results “fit,” or agree, with the results expected based on the null hypothesis. In the case of genetics, we would ask, for example, how well the results of a cross fit with the expectations of a Mendelian hypothesis.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The Chi-Square Method Suppose that we crossed a plant having purple flowers with a plant having white flowers and, among the progeny, observed 14 plants with purple flowers and 6 plants with white flowers. Is this result close enough to be accepted as a 1 : 1 ratio? What if we observed 15 plants with purple flowers and 5 plants with white flowers? Are either of these results consistent with a 1 : 1 ratio? There is bound to be random variation in the observed results from one experiment to the next. Who is to say which results are consistent with a particular genetic hypothesis? A conventional measure of goodness of fit is a value called chisquare (its symbol is χ2), which is calculated from the number of progeny observed in each of various classes, compared with the number expected in each of the classes on the basis of some genetic hypothesis. For example, in a cross between plants with purple flowers and those with white flowers, we may be interested

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

in testing the hypothesis that the parent with purple flowers is heterozygous for one pair of alleles determining flower color and that the parent with white flowers is homozygous recessive. Suppose further that we examine 20 progeny plants from the mating and find that 14 are purple and 6 are white. The procedure for testing this genetic hypothesis (or any other genetic hypothesis) by means of the chi-square method is as follows: 1. State the genetic hypothesis in detail, specifying the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

genotypes and phenotypes of the parents and the possible progeny. In the example using flower color, the genetic hypothesis implies that the genotypes in the cross purple × white could be symbolized as Pp × pp. The possible progeny genotypes are Pp and pp. 2. Use the rules of probability to make explicit predictions of the types and proportions of progeny that should be observed if the genetic hypothesis is true. Convert the proportions to numbers of progeny (percentages are not allowed in a χ2 test). If the hypothesis about the flower-color cross is true, then we should expect the progeny genotypes Pp and pp to occur in a ratio of 1 : 1. Because the hypothesis is that Pp flowers are purple and pp flowers are white, we expect the phenotypes of the progeny to be purple or white in the ratio 1 : 1. Among 20 progeny, the expected numbers are 10 purple and 10 white. 3. For each class of progeny in turn, subtract the expected number from the observed number. Square this difference and divide the result by the expected number. In our example, the calculation for the purple progeny is (14 − 10)2/10 = 1.6, and that for the white progeny is (6 − 10)2/10 = 1.6. 4. Sum the result of the numbers calculated in step 3 for all classes of progeny. The summation is the value of χ2 for

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

these data. The sum for the purple and white classes of progeny is 1.6 + 1.6 = 3.2, which is the value of χ2 for the experiment, calculated on the assumption that our genetic hypothesis is correct. In symbols, the calculation of χ2 can be represented by the expression

in which ∑ means summation over all the classes of progeny. Note that χ2 is calculated using the observed and expected numbers— not the proportions, ratios, or percentages. Using something other than the actual numbers is the most common beginner's mistake in applying the χ2 method. The χ2 value is reasonable as a measure of goodness of fit, because the closer the observed numbers are to the expected numbers, the smaller the value of χ2. A value of χ2 = 0 means that the observed numbers fit the expected numbers perfectly. Note also that χ2 cannot be negative.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

But what constitutes a value of χ2 that is sufficiently large to reject our null hypothesis? At this point, the logic of distributions enters the picture. Similar to the case of the coin flips, we can ask what outcomes we would expect to observe less than 5 percent of the time, given the expectation of the null hypothesis, with the difference being measured by the value of χ2 calculated from the data. Before we can do so, however, we need to consider one more factor—the number of degrees of freedom of the particular χ2 test, which is related to the number of classes of the data. The more classes of data, the more values there are that can deviate from the expected value, and, therefore, the higher the value of χ2 we might observe simply by chance.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

The number of degrees of freedom is usually the number of classes in the data minus . The reason for subtracting 1 is that, in calculating the expected numbers of progeny, we make sure that the total number of progeny is the same as that actually observed. For this reason, one of the classes of data is not really “free” to contain any number we might specify; because the expected number in one class must be adjusted to make the total come out correctly, one “degree of freedom” is lost. Analogous χ2 tests with three classes of data have 2 degrees of freedom, and those with four classes of data have 3 degrees of freedom. In the case described earlier, the number of degrees of freedom is 2 − 1, or 1. With that information and our calculated value of χ2, we can then do a simulation similar to that shown in Figure 4.22, except that we calculate χ2 for 1000 “experiments” with 1 degree of freedom. The results, shown in FIGURE 4.23, indicate that if χ2

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

is greater than the critical value of 3.84, then we would expect to see this much or more deviation from the expected result less than 5 percent of the time.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

FIGURE 4.23 Histogram showing the value of χ2 obtained in 1000

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

simulated experiments with 1 degree of freedom. Outcomes shown in red represent the 5 percent that differ most from the expectations of the null hypothesis (χ2 > 3.84; see Table 4.5).

As another example of the calculation of χ2, suppose that the progeny of an F1 × F1 cross includes two contrasting phenotypes observed in the numbers 99 and 45. In this case, the genetic hypothesis might be that the trait is determined by a pair of alleles of a single gene, in which case the expected ratio of dominant : recessive phenotypes among the F2 progeny is 3 : 1. Considering the data, the question is whether the observed ratio of 99 : 45 is in satisfactory agreement with the expected 3 : 1. Calculation of the value of χ2 is illustrated in TABLE 4.4. The total number of progeny

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

is 99 + 45 = 144. The expected numbers in the two classes, on the basis of the genetic hypothesis that the true ratio is 3 : 1, are calculated as (3/4) × 144 = 108 and (1/4) × 144 = 36. Because there are two classes of data, there are two terms in the χ2 calculation:

TABLE 4.4 Calculation of χ2 for a monohybrid ratio Phenotype

Observed

Expected

Deviation from

(class)

number

number

expected

Wild type

99

108

−9

0.75

Mutant

45

36

+9

2.25

Total

144

144

χ2 = 3.00

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Once the χ2 value has been calculated, the next step is to interpret whether this value represents a good fit or a bad fit to the expected numbers. This assessment is done with the values in TABLE 4.5. In the table, rows are degrees of freedom, and columns are the probabilities P that a worse fit (or one equally bad) would be obtained by chance, assuming that the genetic hypothesis is true. If the genetic hypothesis is true, then the observed numbers should be reasonably close to the expected numbers. If the observed χ2 is so large that the probability of a fit as bad or worse is very small, then the observed results do not fit the theoretical expectations. In such a case, the genetic hypothesis used to calculate the expected numbers of progeny must be rejected, because the observed numbers of progeny deviate too much from the expected numbers.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

TABLE 4.5 Critical values of χ2

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Probability d.f.

0.5

0.1

0.05

0.025

0.01

1

0.45

2.71

3.84

5.02

6.63

2

1.39

4.61

5.99

7.38

9.21

3

2.37

6.25

7.81

9.35

11.34

4

3.36

7.78

9.49

11.14

13.28

5

4.35

9.24

11.07

12.83

15.09

6

5.35

10.64

12.59

14.45

16.81

7

6.35

12.02

14.07

16.01

18.48

8

7.34

13.36

15.51

17.53

20.09

9

8.34

14.68

16.92

19.02

21.67

10

9.34

15.99

18.31

20.48

23.21

In practice, the critical values of P are conventionally chosen as 0.05 (the 5 percent level) and 0.01 (the 1 percent level). For P values ranging from 0.01 to 0.05, the probability that chance alone would lead to a fit as bad or worse is between 1 in 20 experiments and between 1 in 100, respectively. If the P value falls in this range, the correctness of the genetic hypothesis is considered very doubtful, and the result is said to be statistically significant at the 5 percent level. For P values smaller than 0.01, the probability that chance alone would lead to a fit as bad or worse is less than 1 in 100 experiments. In this case, the result is said to be highly significant at the 1 percent level, and the genetic hypothesis is

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

rejected outright. If the terminology of statistical significance seems backward, it is because the term significant refers to the magnitude of the deviation between the observed and expected numbers; in a result that is statistically significant, there is a large (“significant”) difference between what is observed and what is expected. Once we have determined the appropriate number of degrees of freedom, we can interpret the χ2 value in Table 4.5. In the case of our F2 progeny, since there are two classes in the data, there is 2 − 1 = 1 degree of freedom, so we will concentrate on the first row of the table. Remembering that our calculated χ2 value was 3.00, we can compare it to the values in the table. We see that it is less than the value associated with p = 0.05, meaning that chance alone would lead to a fit this bad or worse more than 5 percent of the time. Using the standards described earlier, we would conclude that the results we observed are consistent with our null hypothesis of a 3 : 1 ratio.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

As a second illustration of the χ2 test, we will determine the goodness of fit of Mendel's round-versus-wrinkled data to the expected 3 : 1 ratio. Among the 7324 seeds that Mendel observed, 5474 were round and 1850 were wrinkled. The expected numbers are (3/4) × 7324 = 5493 round and (1/4) × 7324 = 1831 wrinkled. The χ2 value is calculated as follows:

The fact that the χ2 is less than 1 already implies that the fit is very good. To find out how good, note that the number of degrees of freedom equals 2 − 1 = 1 because there are two classes of data (round and wrinkled). From Table 4.5, the P value for χ2 = 0.26

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

with 1 degree of freedom is much greater than 0.05; indeed, it is approximately 0.65. Thus, in approximately 65 percent of all experiments of this type, a fit as bad or worse would be expected simply because of chance. Only about 35 percent of all experiments would yield a better fit.

Are Mendel's Data Too Good to Be True? Many of Mendel's experimental results are very close to the expected values. For the ratios listed in the chapter titled Mendelian Genetics: The Principle of Segregation and Assortment, the χ2 values are 0.26 (round versus wrinkled seeds), 0.01 (yellow versus green seeds), 0.39 (purple versus white flowers), 0.06 (inflated versus constricted pods), 0.45 (green versus yellow pods), 0.35 (axial versus terminal flowers), and 0.61 (long versus short stems). (As an exercise in working with χ2, you should confirm these calculations for yourself.) All of the χ2 tests

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

have P values of 0.45 or greater, which means that the reported results are in excellent agreement with the theoretical expectations. In 1936, the statistician Ronald Fisher pointed out that Mendel's results are suspiciously close to the theoretical expectations. In a large number of experiments, some experiments can be expected to yield fits that appear doubtful simply because of chance variation from one experiment to the next. In Mendel's data, the doubtful values that are to be expected appear to be missing. FIGURE 4.24 shows the observed deviations in Mendel's experiments compared with the deviations expected by chance. (The measure of deviation is the square root of the χ2 value, assigned either a plus or a minus sign according to whether the dominant or the recessive phenotypic class was in excess of the expected number.) For each magnitude of deviation, the height of the bar on the right gives the number of experiments that Mendel observed with such a magnitude of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

deviation; the height of the bar on the left gives the number of experiments expected to deviate by this amount as a result of chance alone. There are clearly too few experiments with deviations smaller than −1 or larger than +1. This type of discrepancy could be explained if Mendel discarded or repeated a

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

few experiments with large deviations that made him suspect that the results were not to be trusted.

FIGURE 4.24 Distribution of deviations observed in 69 of Mendel's experiments (yellow bars) compared with the expected values (orange bars). There is no suggestion that the data in the middle have been adjusted to improve the fit, but there are fewer experiments with large deviations than might be expected. Several experiments with large deviations may have been discarded or repeated.

Did Mendel cheat? Did he deliberately falsify his data to make them appear better? Mendel's paper reports extremely deviant ratios from individual plants, as well as experiments repeated when

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

the first results were doubtful. These are not the kinds of things a dishonest person would admit. Only a small bias is necessary to explain the excessive goodness of fit in Figure 4.24. In a count of seeds or individual plants, only about 2 phenotypes per 1000 would need to be assigned to the wrong category to account for the bias in the 91 percent of the data generated by the testing of monohybrid ratios. The excessive fit could also be explained if three or four entire experiments were discarded or repeated because deviant results were attributed to pollen contamination or other accident. After careful reexamination of Mendel's data in 1966, the evolutionary geneticist Sewall Wright concluded:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Mendel was the first to count segregants at all. It is rather too much to expect that he would be aware of the precautions now known to be necessary for completely objective data…. Checking of counts that one does not like, but not of others, can lead to systematic bias toward agreement. I doubt whether there are many geneticists even now whose data, if extensive, would stand up wholly satisfactorily under the χ2 text…. Taking everything into account, I am confident that there was no deliberate effort at falsification. Mendel's data are some of the earliest and most complete “raw data” ever published in genetics (in recent years, publication of raw data has become a common practice). Additional examinations of the data will surely be carried out as new statistical approaches are developed. However, the principal point to be emphasized is that up to the present time, no reputable statistician has alleged

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

that Mendel knowingly and deliberately adjusted his data in favor of the theoretical expectation.

SUMMING UP ■ A test for goodness of fit determines whether the differences between a set of observations and the results expected based on a null hypothesis can be explained by chance. ■ The χ2 test provides a means of testing for goodness of fit to a genetic hypothesis. ■ The probability of observing a particular value of χ2 is dependent on the number of degrees of freedom in the data. ■ Departures from a null hypothesis are statistically significant if the probability of observing the calculated χ2 value as a result of chance is less than 5 percent and highly significant if it is less than 1 percent.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Mendel's data show an exceptional fit to the hypothesis of random segregation, but it is unlikely that this outcome resulted from any deliberate misrepresentation of his results.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

CHAPTER SUMMARY ■ Chromosomes in eukaryotic cells are usually present in pairs. ■ The chromosomes of each pair separate in meiosis, with one chromosome going to each gamete. ■ In meiosis, the chromosomes of different pairs undergo independent assortment because nonhomologous chromosomes move independently. ■ In many animals, sex is determined by a special pair of chromosomes, X and Y. ■ The “crisscross” pattern of inheritance of X-linked genes is determined by the fact that a male receives his X chromosome only from his mother and transmits it only to his daughters. ■ Irregularities in the inheritance of an X-linked gene in Drosophila gave experimental proof of the chromosomal theory of heredity. ■ The progeny of genetic crosses follow the binomial probability formula.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The chi-square statistical test is used to determine how well the observed genetic data agree with the expectations derived from a hypothesis.

REVIEW THE BASICS ■ What is the genetic significance of the fact that gametes contain half the chromosome complement of somatic cells? ■ The term mitosis derives from the Greek mitos, which means “thread.” The term meiosis derives from the Greek meioun, which means “to make smaller.” Which feature, or features, of these types of nuclear division might have led to the choice of these terms?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

■ Explain the meaning of the terms reductional division and equational division. What is reduced or kept equal? To which nuclear divisions do the terms refer? ■ Explain the following statement: “Independent alignment of nonhomologous chromosomes at metaphase I of meiosis is the physical basis of independent assortment of genes in different chromosomes.” ■ What are some of the important differences between the first meiotic division and the second meiotic division? ■ Draw a diagram of a bivalent and label the following parts: centromere, sister chromatids, nonsister chromatids, homologous chromosomes, chiasma. ■ What are the principal characteristics of human pedigrees in which a rare X-linked recessive allele is segregating? ■ In which ways is the inheritance of a Y-linked gene different from that of an X-linked gene? ■ How did nondisjunction “prove” the chromosome theory of heredity? ■ Why is a statistical test necessary to determine whether an observed set of data yields an acceptable fit to the result expected from a particular genetic hypothesis?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Which statistical test is often used for this purpose? ■ What are the conventional P values for significant and highly significant, and what do these numbers mean?

GUIDE TO PROBLEM SOLVING PROBLEM 1 A recessive mutation in an X-linked gene results in hemophilia, marked by a prolonged increase in the time needed for blood to clot. Suppose that two phenotypically normal parents

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

produce three normal daughters and a son affected with hemophilia. (a) What is the probability that all of the daughters are heterozygous carriers? (b) If one of the daughters mates with a normal male and produces a son, what is the probability that the son will be affected? ANSWER (a) Because the phenotypically normal parents have an affected son who gets his only X chromosome from his mother, the mother must be a carrier of the mutation. Therefore, the probability that any particular daughter is a carrier is ½. The probability that all three daughters are carriers is (½)3 because their births are independent events.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(b) If the daughter is not a carrier, the probability of an affected son is 0; if the daughter is a carrier, the probability of an affected son is ½. Because the probability of the daughter being a carrier is ½ [from part (a)], the overall probability of an affected son is (½) × 0 + (½) × (½) = ¼. PROBLEM 2 In a sibship with seven children, what is the probability that it includes four boys and three girls or three boys and four girls? Assume that each child has an equal likelihood of being a boy or a girl. ANSWER The probability that the sibship consists of four boys and three girls, in any order of birth, equals [7!/(4!3!)](½)4(½)3, where the factor in square brackets is the number of possible birth orders of four boys and three girls. This probability works out to 35/128.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Likewise, the probability that the sibship consists of three boys and four girls is [7!/(3!4!)](½)3(½)4, which also equals 35⁄128. The question asks for the probability of either four boys and three girls or three boys and four girls, and these events are mutually exclusive; hence the overall probability is 35/128 + 35/128 = 35/64, or about 55 percent. PROBLEM 3 A geneticist carries out a cross between two strains of fruit flies that are heterozygous for a recessive allele of each of two genes, st (scarlet) and bw (brown), affecting eye color. Flies homozygous for st have bright red (scarlet) eyes, whereas those with the st+/st+ and st+/st genotypes have wild-type brick-red eyes. Flies homozygous for bw have brown eyes, whereas those with the bw+/bw+ and bw+/b genotypes have wild-type brick-red eyes. The double homozygous genotype st/st, bw/bw results in white eyes. In a test of independent assortment of these two genes, a geneticist crosses st+/st, bw+/bw females with st+/st, bw+/bw males. Among 240 progeny there are 150 flies with wild-type eyes, 36 with scarlet eyes, 46 with brown eyes, and 8 with white eyes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(a) Under the null hypothesis that these genes undergo independent assortment, what are the expected numbers in each phenotypic class? (b) What is the value of chi-square in a test of goodness of fit between the observed values and the expected values based on the null hypothesis of independent assortment? (c) How many degrees of freedom does this chi-square value have? (d) What is the P-value for the chi-square value in this goodness-of-fit test? Does this P-value support the null hypothesis of independent assortment, or should the hypothesis be rejected?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

(e) Would a greater chi-square value increase or decrease the P-value? ANSWER (a) Because this is a dihybrid cross st+/st, bw+/bw × st+/st, bw+/bw, the ratio of 9 : 3 : 3 : 1 of the offspring phenotypes should be expected if the genes assort independently. We can calculate the expected number of flies in each class of the progeny: Phenotype

Progeny

Expected number

wild type

st+/−, bw+/−

(9/16) × 240 = 135

scarlet

st/st, bw+/−

(3/16) × 240 = 45

brown

st+/−, bw/bw

(3/16) × 240 = 45

white

st/st, bw/bw

(1/16) × 240 = 15

(b) The chi-square value is calculated as follows:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

where the sum is over all classes of progeny. In this case

(c) The number of degrees of freedom equals the number of classes of data minus 1. In this case, there are four classes of data; thus there are 3 degrees of freedom. (d) The P-value for a chi-square value of 6.76 with 3 degrees of freedom equals 0.08. This P-value is greater than 0.05.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Therefore, we should not reject the null hypothesis of independent assortment. (e) A greater chi-square value would decrease the P-value. PROBLEM 4 The mutation for bar-shaped eyes in Drosophila melanogaster shows the following features of genetic transmission: (a) The mating of bar-eyed males with wild-type females produces wild-type sons and bar-eyed daughters. (b) The bar-eyed females from the mating in (a), when mated with wild-type males, yield a 1 : 1 ratio of bar : wild-type sons and a 1 : 1 ratio of bar : wild-type daughters.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Which mode of inheritance do these characteristics suggest? ANSWER Because the sexes are affected unequally in the progeny of the mating in (a), some association with the sex chromosomes is suggested. The progeny from mating (a) provide the important clue. Because a male receives his X chromosome from his mother, the observation that all males are wild type suggests that the gene for bar eyes is on the X chromosome. The fact that all daughters are affected is also consistent with X-linkage, provided that the bar mutation is dominant. Mating (b) confirms the hypothesis of a dominant X-linked gene, because the females from mating (a) would have the genotype bar/+ and so would produce the observed progeny. Thus, the data are consistent with the bar mutation being an X-linked dominant.

ANALYSIS AND APPLICATIONS 4.1 The accompanying diagrams illustrate a pair of homologous chromosomes in prophase I of meiosis. Which diagram corresponds to each stage:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

leptotene, zygotene, pachytene, diplotene, and diakinesis?

4.2 The accompanying diagrams illustrate anaphase in an organism that has two pairs of homologous chromosomes. Identify the stages as mitosis, meiosis I, or meiosis II.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

4.3 What is the probability that four offspring from the mating Aa × Aa consist of exactly 3 A— and 1 aa? 4.4 A woman who is heterozygous for both a phenylketonuria mutation and an X-linked hemophilia mutation has a child with a phenotypically normal man who is also heterozygous for a phenylketonuria mutation. What is the probability that the child will be affected by both diseases? (Assume that the couple is equally likely to have a boy or a girl.) 4.5 In the accompanying pedigree, the purple symbols represent individuals affected with X-linked

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Duchenne muscular dystrophy. What is the probability that the woman III-3 is a heterozygous carrier?

4.6 A chromosomally normal woman and a chromosomally normal man have a son whose sexchromosome constitution is XYY. In which parent, and in which meiotic division, did the nondisjunction take place? 4.7 Drosophila virilis is a diploid organism with 6 pairs of chromosomes (12 chromosomes altogether). How many chromatids and chromosomes are present in the following stages of cell division: (a) Metaphase of mitosis? (b) Metaphase I of meiosis? Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(c) Metaphase II of meiosis?

4.8 In a trihybrid cross with genes that undergo independent assortment: (a) What is the expected proportion of triply heterozygous offspring in the F2 generation? (b) What is the expected proportion of triply homozygous offspring in the F2 generation?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.9 The diploid gekkonid lizard Gonatodes taniae from Venezuela has a somatic chromosome number of 16. If the centromeres of the 8 homologous pairs are designated as Aa, Bb, Cc, Dd, Ee, Ff, Gg, and Hh: (a) How many different combinations of centromeres could be produced during meiosis? (b) What is the probability that a gamete will contain only those centromeres designated by capital letters?

4.10 Among sibships consisting of six children, and assuming a sex ratio of 1 : 1: (a) What is the proportion with no girls? (b) What is the proportion with exactly one girl? (c) What is the proportion with exactly two girls? (d) What is the proportion with exactly three girls?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(e) What is the proportion with three or more boys?

4.11 A litter of cats includes eight kittens. What is the probability that it contains even numbers of both males and females? Assume an equal likelihood of male and female kittens, and for purposes of this problem regard 0 as an even number. Does the answer surprise you? Why? 4.12 A horticulturalist crossed a true-breeding onion plant with red bulbs and a true-breeding onion plant with white bulbs. All of the F1 plants had white bulbs. When seeds resulting from self-fertilization of the F1 plants were grown, onion bulbs were recovered in the ratio of 12 white bulbs : 3 red bulbs : 1 yellow bulb. Propose a hypothesis to explain these observations.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.13 Among 160 progeny in the F2 generation of a dihybrid cross, a geneticist observes four distinct phenotypes in the ratio 91: 21 : 37 : 11. She believes this result may be consistent with a ratio of 9 : 3 : 3 : 1. To test this hypothesis, she calculates the chi-square value. Does the test support her hypothesis, or should she reject it? 4.14 What is the value of the chi-square that tests goodness of fit between the observed numbers 40 : 60 as compared with the expected numbers 50 : 50? 4.15 What is the probability that a sibship of seven children includes at least one boy and at least one girl? Assume that the sex ratio is 1 : 1.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

4.16 How many genotypes are possible for an autosomal gene with six alleles? How many genotypes are possible with an X-linked gene with six alleles? 4.17 The growth habit of the Virginia groundnut Arachis hypogaea can be “runner” (spreading) or “bunch” (compact). Two pure-breeding strains of groundnuts with the bunch growth habit are crossed. The F1 plants have the runner growth habit. If the plants are allowed to self-fertilize, the F2 progeny ratio is 9 runner : 7 bunch. Which genetic hypothesis can account for these observations? 4.18 In some human pedigrees, the blue/brown eye color variation segregates like a single-gene difference, with the brown allele being dominant relative to the blue allele. Two brown-eyed individuals, each of whom had a blue-eyed parent, mate. Assuming that

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

the trait segregates like a single-gene difference in this pedigree: (a) What is the probability that the first child will have brown eyes? (b) If a brown-eyed child results, what is the probability that the child will be heterozygous? (c) What is the probability that this couple will have three children with blue eyes? That none of the three children will have blue eyes?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

4.19 In the accompanying pedigree, the male I-2 is affected with red-green color blindness owing to an X-linked mutation. What is the probability that male IV-1 is color blind? (Assume that the only possible source of the color blindness mutation in the pedigree is that from male I-2.)

4.20 The male II-1 in the accompanying pedigree is affected with a trait due to a rare X-linked recessive allele. Use the principles of conditional probability to calculate the probability that the female III-1 is a heterozygous carrier.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

4.21 The accompanying pedigree is the same as that in Problem 4.20 but with additional information based on molecular analysis. The male II-1 is again affected with a trait due to a rare X-linked recessive allele. The bands in the gel are PCR products of an SSRP that identified the wild-type allele (the shorter allele S corresponding to the band near the bottom of the gel) or the mutant allele (the longer allele L corresponding to the band near the top of the gel). From the information given in the pedigree and the gels, calculate the probability that the female III-1 is a heterozygous carrier.

4.22 In the accompanying pedigree, the males I-2 and III-2 are affected with a trait due to a common X-linked recessive allele. What is the probability that individual III-1 is heterozygous?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.23 Yellow body color in Drosophila is determined by the recessive allele y of an X-linked gene, and the wildtype gray body color is determined by the y+ allele. What genotype and phenotype ratios would be expected from the following crosses? (a) Yellow male × wild-type female (b) Yellow female × wild-type male (c) Daughter from mating in part (a) × wild-type male

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(d) Daughter from mating in part (a) × yellow male

4.24 The accompanying pedigree and gel diagram show the molecular phenotypes for an SSRP with two alleles yielding PCR products of the indicated sizes in base pairs. Which mode of inheritance does the pedigree suggest? Based on this hypothesis, and using A1 to represent the SSRP allele associated with the 3-kb band and A2 to represent that associated with the 8-kb band, deduce the genotype of each individual in the pedigree.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

4.25 Tall, yellow-flowered crocus is mated with short, white-flowered crocus. Both varieties are truebreeding. All of the F1 plants are backcrossed with the short, white-flowered variety. This backcross yields 800 progeny in the following proportions: 234 tall yellow, 203 tall white, 175 short yellow, and 188 short white plants. Does the observed result fit the genetic hypothesis of 1 : 1 : 1 : 1 segregation as assessed by a χ2 test? 4.26 Attached-X chromosomes in Drosophila are formed from two X chromosomes attached to a common centromere. Females of genotype C(1)RM/Y, in which C(1)RM denotes the attached-X chromosomes, produce C(1)RM-bearing and Ybearing gametes in equal proportions. What progeny is expected to result from the mating between a male carrying the X-linked allele, y, for yellow body and an attached-X female with wild-type body? How does this result differ from the typical pattern of X-linked inheritance? (Note: Drosophila zygotes containing three X chromosomes or no X chromosomes do not survive.)

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

4.27 A rare dominant autosomal mutation W results in wooly, curled hair in some European pedigrees. A woman with wooly hair with blood group O marries a man with straight hair (ww) with blood group AB. The genes are in different chromosomes. (a) What are the chances that the mating will produce a wooly-haired, group B child? (b) What are the chances that the mating will produce a straight-haired, group B child? (c) If three straight-haired children with blood group A are born to these parents, what are the chances that the next child born will be wooly-haired and blood group B?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

4.28 Ducklings of the domesticated mallard Anas platyrhynchos may exhibit dark brown dorsal plumage known as mallard, almost black dorsal plumage known as dusky, or a pattern known as restricted in which the black is confined to patches on the head and tail. These phenotypes result from the action of three alleles of a single autosomal gene. Three types of crosses were made, with the following results: 1. Restricted × mallard: All F1 are restricted; crosses of F1 × F1 yield an F2 generation with the ratio 3 restricted : 1 mallard. 2. Mallard × dusky: All F1 are mallard; crosses of F1 × F1 produce an F2 generation with the ratio 3 mallard : 1 dusky. 3. Restricted × dusky: All F1 are restricted; crosses of F1 × F1 produce an F2 generation with the ratio 3 restricted : 1

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

dusky. (a) Assume that an F1 male from cross 1 is mated with an F1 female from cross 2. List the phenotypes and their expected frequencies among the progeny of this cross. (b) Assume that an F1 male from cross 3 is mated with an F1 female from cross 2. List the phenotypes and their expected frequencies among the progeny of this cross.

4.29 In cattle, an allele for the absence of horns shows complete dominance: HH and Hh are hornless or polled, and hh is horned. In contrast, the effect of the allele producing red coat (R) shows incomplete dominance with the allele producing white coat color (r). The heterozygous genotype Rr is roan colored— an intermediate color in which white hairs are intermixed with red hairs. H and R undergo independent assortment. (a) What would be the phenotype of an F1 offspring from the mating RR HH × rr hh?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(b) What would be the phenotypes and their expected proportions in the F2 progeny of the cross F1 × F1 from part (a)? (c) What would be the phenotypes and their expected proportions among the progeny derived from crossing F1 individuals from part (a) to the original horned white stock?

4.30 Familial Mediterranean Fever (FMF) is a hereditary inflammatory disorder that affects groups of patients

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

originating from around the Mediterranean Sea, especially Armenian and non-Ashkenazi Jewish populations, Anatolian Turks, and Levantine Arabs. FMF is inherited in an autosomal recessive pattern. It is not known whether the gene causing the disease is the same in these different populations. To investigate this possibility, linkage of the allele to a series of SSRPs was examined. One SSRP in chromosome 16 was especially informative. For this SSRP, two extended Armenian families from the same geographic region gave the results shown in the accompanying pedigrees and gels. The green symbols represent affected individuals. The numbers between the gels correspond to the SSRP alleles. (a) Is there evidence for linkage of the allele for FMF to any SSRP allele?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(b) Do any of the data conflict with your explanation? If so, how do you account for it?

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 The accompanying pedigree and gel diagram pertain to a morphological phenotype (green symbols in the pedigree) and a molecular phenotype (a two-allele SSRP; sizes

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

of PCR products in base pairs are shown). Which mode of inheritance do these data suggest for each trait?

CHALLENGE PROBLEM 2 In mice, the dominant T allele results in a short tail, but the homozygous T/T genotype is lethal. A cross between two short-tailed mice produces a litter of five pups. What is the expected distribution of tailless to tailed mice? Why are the two most likely outcomes equally probable?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

CHALLENGE PROBLEM 3 X-linked hemophilia is present in the accompanying pedigree, as indicated by red symbols. What is the probability that the woman III-5 is a carrier, taking into account the information that she has had two normal sons? What is the probability that her next son will be affected?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

FOR FURTHER READING Fisher, R. A. (1936). Has Mendel's work been rediscovered? Annals of Science, 1(2), 115–137. http://doi.org/10.1080/00033793600200111 The eminent geneticist and statistician R. A. Fisher addresses the question of whether Mendel's ratios were “too good.” Salz, H. (2016). Sex determination in Drosophila: The view from the top. Fly, 4(1), 60–70. http://doi.org/10.4161/fly.4.1.11277 A somewhat technical but very thorough overview of the sex determination pathway in Drosophila. Sutton, W. S. (1902). On the morphology of the chromosome group in Brachystola magna. Biological Bulletin, 4, 24–39.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Sutton, W. S. (1903). The chromosomes in heredity. Biological Bulletin, 4, 231–251. Two papers by Walter Sutton providing some of the earliest experimental evidence of a link between chromosomes and heredity.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:30.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Martin Shields/Alamy Stock Photos.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

CHAPTER 5 Genetic Linkage and Chromosome Mapping CHAPTER OUTLINE 5.1 Linkage and Recombination of Genes in a Chromosome

5.2 Genetic Mapping 5.3 Genetic Mapping in a Three-Point Testcross 5.4 Mapping by Tetrad Analysis 5.5 Genetic Mapping in Humans 5.6 Special Features of Recombination ROOTS OF DISCOVERY Genes All in a Row

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Alfred H. Sturtevant (1913) The Linear Arrangement of Six Sex-Linked Factors in Drosophila, as Shown by Their Mode of Association ROOTS OF DISCOVERY Mapping Markers in the Human Genome David Botsein, Raymond White, Mark Skolnick, and Ronald Davis (1980) Construction of a Genetic Linkage Map in Man Using Restriction Fragment Length Polymorphisms

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

LEARNING OBJECTIVES & SCIENCE COMPETENCIES The principles of genetic linkage and chromosome mapping are essential in the discovery of disease genes and genetic risk factors for disease. When you understand these principles, you will be able to carry out the following science competencies: ■ Given a genetic map, predict the kinds and relative frequencies of gametes that would be produced by an individual of a specified genotype. ■ Analyze the results of a genetic cross with three linked genes to deduce the genotypes of the parents, the order of the genes along the chromosome, the map distances between the genes, and the degree of interference between crossovers.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Detect linkage between a gene and its centromere in an organism with ordered tetrads, and estimate the map distance between the gene and its centromere. ■ Calculate and interpret lod scores based on simple pedigrees. ■ Interpret the results of a genome-wide association scan. ecall from the Medelian Genetics: The Principles of Segregation and Assortment chapter that Mendel examined seven traits in garden peas and was able to establish that they all assort independently. However, we saw in the chapter The Chromosomal Basis of Inheritance that chromosomes, as opposed

R

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

to individual genes, are the fundamental units of segregation and assortment. Mendel, of course, did not know that, but subsequent researchers have been able to extend the genetic analysis of garden peas. We now know that they contain seven pairs of autosomes; furthermore, the genes characterized by Mendel have been localized on those chromosomes (TABLE 5.1).

TABLE 5.1 Mendel's genes Trait

Cotyledon

Dominant

Recessive

Gene function

Chromosome

phenotype

phenotype

Yellow

Green

Stay-green gene

I

Purple

White

bHLH transcription

II

color Seed coat/flower

factor

color Stem length

Tall

Dwarf

GA 3-oxidase1

III

Pod form

Inflated

Constricted

Sclerenchyma

III

formation in pods Position of

Axial

Terminal

Meristem function

IV

Round

Wrinkled

Starch branching

V

flowers

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Seed shape

enzyme 1 Pod color

Green

Yellow

Chloroplast

V

structure in pod wall

At first glance, these results appear a bit puzzling. Because homologous chromosomes behave as units when they undergo meiosis, one might expect that genes located in the same

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

chromosome would not undergo the same pattern of independent assortment observed in nonhomologous chromosomes. Therefore, how is it that genes on the same chromosome (such as those determining seed shape and pod color, both on chromosome V) underwent independent assortment in Mendel's experiments? To understand that outcome, we now turn to the subject of linkage and its applications to genetic mapping. Genes that are always transmitted together are said to show

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

complete linkage. Pioneering studies of this issue were carried out by Thomas Hunt Morgan, who studied mutations located in the X chromosome of Drosophila. He did observe linkage, but what he observed was an incomplete form of linkage, evidenced by progeny who exhibited a combination of traits not exhibited in the parents. Linkage between genes is usually incomplete because homologous chromosomes can undergo an exchange of segments when they are paired. Thus, while the alleles present in each chromosome do tend to remain together in inheritance, some chromosomes can be transmitted that have new combinations of alleles. An exchange event (crossing over) between homologous chromosomes results in the recombination of genes, yielding daughter chromosomes that carry combinations of alleles not present in the parental chromosomes. What we will see in this chapter is that the probability of crossing over between any two genes serves as a measure of genetic distance between the genes. This makes it possible to construct a genetic map—a diagram of a chromosome showing the relative positions of the genes. The genetic mapping of linked genes is an important research tool in genetics because it enables a new gene to be assigned to a chromosome, and even to a precise position relative to other genes in the same chromosome. Genetic mapping is often a first step in

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

the identification and isolation of a new gene. It is essential in

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

human genetics for the identification of genes associated with hereditary diseases, such as the genes whose presence predisposes women carriers to the development of breast cancer.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

5.1 Linkage and Recombination of Genes in a Chromosome As we saw in the chapter titled The Chromosomal Basis of Inheritance, a direct test of independent assortment is to carry out a testcross between an F1 double heterozygote (Aa Bb) and the double recessive homozygote (aa bb). Shown in FIGURE 5.1 are

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the expected gametes from the Aa Bb parent when the genes are in different chromosomes. Because the possible metaphase I alignments of homologous chromosomes are equally likely, the double heterozygote produces all four possible types of gametes in equal proportions. The alleles present in a gamete are typically written in a row with a space between each gene; in this case, the four equally likely types of gametes are A B, A b, a B, and a b.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

FIGURE 5.1 The physical basis of the independent assortment of genes in different chromosomes is that nonhomologous chromosomes orient independently at metaphase I of meiosis. In this example, the orientation on the left results in A B and a b gametes, whereas that on the right results in A b and a B gametes.

In Morgan's early experiments with Drosophila, he examined the segregation of the X-linked alleles w versus w+, which determine Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

white versus normal red eyes, and the alleles m versus m+, which determine miniature versus normal wing size. When genes are linked, it is important to keep track of which alleles are present together in the same chromosome. This is generally done by using a slash (also called a virgule) to separate the alleles present in homologous chromosomes. Morgan's initial cross was between females with white eyes and normal wings and males with red eyes and miniature wings. Using the slash, we would write this cross as follows:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

As expected, the resulting F1 progeny consisted of wild-type females (genotype w m+/w+m) and white-eyed, nonminiature males (genotype w m+/Y). Then the F1 progeny were crossed:

In this cross, X chromosomes carrying either the allele combination w m+ or the allele combination w+m are called parental types of chromosomes; this terminology reflects the fact that parents involved in the cross carry the alleles in these configurations. On the one hand, if the w and m genes were completely linked, then all of the offspring would inherit one or the other of the parental types of chromosome. On the other hand, if recombination between the w and m genes took place, then some of the progeny would inherit an X chromosome with either the allele combination w+m+ or the allele combination w m. These are known as recombinant types of chromosomes because they contain allele combinations that differ from those present in the parental chromosomes. When the cross displayed previously was carried out, the female progeny were all nonminiature with a 1 : 1 ratio of red : white eyes

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(if you do not understand why, write out the Punnett square), and the male progeny were as follows:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Because each male receives his only X chromosome from his mother, a male's phenotype reveals the genotype of the X chromosome that he inherited. The results of the experiment show a significant departure from the 1 : 1 : 1 : 1 ratio of the four male phenotypes expected with independent assortment, yet the ratios of w : w+ and of m : m+ are both approximately 1 : 1. This type of deviation from independent assortment indicates that the parental combinations of the alleles for the w and m genes tend to remain together in inheritance, but they are not completely linked. In this experiment, the combinations of alleles in the parental chromosomes were present in 428/644 = 66.5 percent of the F2 males, and recombinant (nonparental) chromosomes were present in 216/644 = 33.5 percent of the F2 males. The 33.5 percent value for the recombinant chromosomes is called the frequency of recombination; it can also be written as 0.335. This frequency of recombination should be contrasted with both the 50 percent recombination expected with independent assortment and the 0 percent recombination expected with complete linkage. The recombinant X chromosomes w+m+ and w m result from

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

crossing over in meiosis in F1 females. In this example, the frequency of recombination between the linked w and m genes is 33.5 percent, but with other pairs of linked genes it ranges from near 0 to 50 percent. Even genes in the same chromosome, if they are sufficiently far apart, can undergo independent assortment, which means that their frequency of recombination equals 50 percent. This implies the following principle:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Genes with recombination frequencies less than 50 percent are present in the same chromosome (linked). Two genes that undergo independent assortment, indicated by a recombination frequency equal to 50 percent, either are in nonhomologous chromosomes or are located far apart in a single chromosome. The possibility that two genes in the same chromosome may be unlinked creates a problem in terminology. We cannot say that all genes in the same chromosome are “linked,” because linkage means a recombination frequency of less than 50 percent, and genes that are sufficiently far apart are actually unlinked. The term that is used instead is synteny: Two genes are said to be syntenic if they are in the same chromosome, regardless of whether they show independent assortment or linkage. Looking back at Table 5.1, this can explain why all of Mendel's traits assorted independently. Those pairs of genes (for example, those affecting seed shape and pod color) that assort independently are indeed syntenic, but the recombination frequency between them is 50 percent.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Coupling Versus Repulsion of Syntenic Alleles Geneticists use a slash notation for linked genes that has the general form w+m/w m+. This notation is a simplification of the following more descriptive but cumbersome form:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

In this form of notation, the horizontal line replaces the slash in separating the two homologous chromosomes. For consistency, the linked genes are always written in the same linear order. This convention makes it possible to indicate the wild-type allele of a gene with a plus sign in the appropriate position. For example, the genotype w m+/w+m could also be written without ambiguity as w +/+ m. This system of gene notation is used for Drosophila and a number of other organisms. A genotype that is heterozygous for each of two linked genes can have the alleles in either of two possible configurations, as shown in FIGURE 5.2 for the w and m genes. In one configuration, called the trans, or repulsion, configuration, the mutant alleles are in opposite chromosomes, and the genotype is written as w +/+ m (Figure 5.2A). In the alternative configuration, called the cis, or coupling, configuration, the mutant alleles are present in the same chromosome, and the genotype is written as w m/+ +.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.2 Possible configurations of the mutant alleles in a genotype heterozygous for two mutations. (A) The trans, or repulsion, configuration has the mutant alleles on opposite chromosomes. (B) The cis, or coupling, configuration has the mutant alleles on the same chromosome.

Morgan's study of linkage between the white and miniature alleles began with the trans configuration. He also studied progeny from

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

the cis configuration of the w and m alleles, which results from the cross of white, miniature females with red, nonminiature males. This cross is shown in the “+” notation as follows:

In this case, the F1 females were phenotypically wild-type double heterozygotes, and the males had white eyes and miniature wings. A cross of the F1 progeny is written, again using the “+” notation, as follows:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

This cross produced the following progeny:

This result demonstrates that the frequency of recombination with w and m in the cis (coupling) configuration is the same as that observed with w and m in the trans (repulsion) configuration—37.7 percent versus 33.5 percent, respectively. The difference is within the range expected from sampling variation among experiments. However, in this case, the chromosomes constituting the parental and recombinant classes of offspring are reversed. They are reversed because the original parents of the F1 female were different: repulsion (w +/+ m) in one case and coupling (w m/+ +) in

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

the other. The repeated finding of equal recombination frequencies in experiments of this kind leads to the following conclusion: Recombination between linked genes takes place with the same frequency whether the alleles of the genes are in the trans (repulsion) configuration or in the cis (coupling) configuration; it is the same no matter how the alleles are arranged.

The Chi-Square Test for Linkage

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Any estimated value of the frequency of recombination (r) is subject to sampling variation from one experiment to the next. How can the statistical significance of an observed value of r < 1/2 be tested against the alternative hypothesis that r = 0.5? To take a specific example, how can we be sure that the value r = 0.377 observed in the test of w m in the coupling configuration is not a statistical accident? Perhaps the true value is r = 0.5 and the genes, although syntenic, are nevertheless unlinked. Statistical significance for linkage can be tested on the basis of the observed numbers by means of the chi-square test examined in the The Chromosomal Basis of Inheritance chapter. As an example, we will use the w m experiment yielding the estimate r = 0.377. The overall chi-square value in the testcross data is given in the top row of TABLE 5.2. With four classes of progeny, there is 4 − 1 = 3 degrees of freedom. Only one of these classes is relevant to the hypothesis of linkage; the others pertain to the segregation of w versus w+ and of m versus m+. This point is shown by the chi-square values in the bottom rows of the table. The test of 1 : 1 segregation of w versus w+ entails pooling the m and m+ data to obtain the observed ratio:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

TABLE 5.2 Testing for linkage with chi-square Chi-square test

Calculation

Degrees

P

of

value

freedom Overall chi square

(395 − 311.75)2/311.75 + (382 −

76.775

3

311.75)2/311.75 + (223 −

1.5 × 10−16

311.75)2/311.75 + (247 − 311.75)2/311.75 = Segregation of w

(605 − 623.5)2/623.5 + (642 −

versus w+

623.5)2/623.5 =

Segregation of m

(629 − 623.5)2/623.5 + (618 −

versus m +

623.5)2/623.5 =

Linkage

(470 − 623.5)2/623.5 + (777 −

(recombinant

623.5)2/623.5 =

1.098

1

0.295

0.097

1

0.755

75.580

1

3.5 × 10−18

versus nonrecombinant) Sum of 1-d.f. chi-

76.775

3

square values

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

To test for 1 : 1 of the m versus m+ segregation, we pool the w and w+ data:

The hypothesis of no linkage (r = 1/2) predicts a 1 : 1 ratio of recombinant : nonrecombinant progeny, so in this case we pool the data as follows:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

The three individual chi-square values, each with 1 degree of freedom (d.f.), are shown in Table 5.2. The overall chi-square with 3 d.f. has a P value that can be calculated by numerical methods as 1.5 × 10−16; this leaves no doubt that the underlying ratio of the four classes of progeny differs from 1 : 1 : 1 : 1. To learn the source of the discrepancy, we examine the individual chi-square values that have 1 d.f. Neither the w versus w+ segregation nor the m versus m+ segregation gives any evidence for departure from 1 : 1, with the P values being 0.295 and 0.755, respectively. However, the chi-square value for recombinants versus nonrecombinants is highly significant, with a P value that can be calculated as 3.5 × 10−18. From this test, we can safely conclude that the cause of the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

discrepancy is linkage, and we have already estimated the frequency of recombination as r = 470/1247 = 0.377. Note also in Table 5.2 that the 1-d.f. chi-square values sum to the overall chi-square value of 76.775. The additivity means that the three hypotheses being tested are independent in the sense that the truth or falsehood of any of them has no bearing on the validity of the others. In practice, we are not often interested in such a detailed analysis of segregation as is outlined in Table 5.2. Usually we want to test only for linkage, and this test is based on the chisquare value for recombinants versus nonrecombinants: The chi-square test for linkage is a test for an equal number (1 : 1 ratio) of recombinant versus nonrecombinant progeny, which has 1 degree of freedom. Testing the individual 1 : 1 segregations is usually unnecessary.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Each Pair of Linked Genes Has a Characteristic Frequency of Recombination The recessive allele y of another X-linked gene in Drosophila results in yellow body color instead of the usual gray color determined by the y+ allele. In one experiment, white-eyed females were mated with males having yellow bodies, and then the wildtype F1 females were testcrossed with yellow-bodied, white-eyed males:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The progeny were

In a second experiment, yellow-bodied, white-eyed females were crossed with wild-type males, and then the F1 wild-type females and F1 yellow-bodied, white-eyed males were intercrossed:

In this case, 98.6 percent of the F2 progeny had parental chromosomes and 1.3 percent had recombinant chromosomes. The parental and recombinant types were reversed in the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

reciprocal crosses, but the recombination frequency was virtually the same. Females with the trans genotype y +/+ w produced about 1.4 percent recombinant progeny, carrying either of the recombinant chromosomes y w or + +; similarly, females with the cis genotype y w/+ + produced about 1.3 percent recombinant progeny, carrying either of the recombinant chromosomes y + or + w. However, the recombination frequency between the genes for yellow body and white eyes was much lower than that between the genes for white eyes and miniature wings (1.4 percent versus about 35 percent). These and other experiments reinforce the following conclusions: ■ The recombination frequency is a characteristic of a particular pair of genes. ■ Recombination frequencies are the same in cis (coupling) and trans (repulsion) heterozygotes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Recombination in Females Versus Males Drosophila is unusual in that recombination does not take place in males. Although it is not known how (or why) crossing over is prevented in males, the absence of recombination in Drosophila males means that all syntenic genes (those located in the same chromosome) show complete linkage in the male. For example, the genes cn (cinnabar eyes) and bw (brown eyes) are both in chromosome 2 but are so far apart that, in females, there is 50 percent recombination. Thus, the cross

yields progeny of genotype + +/cn bw and cn bw/cn bw (the nonrecombinant types) as well as cn +/cn bw and + bw/cn bw (the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

recombinant types) in the proportions 1 : 1 : 1 : 1. However, because there is no crossing over in males, the reciprocal cross

yields progeny only of the nonrecombinant genotypes ++/cn bw and cn bw/cn bw in equal proportions. The absence of recombination in Drosophila males is a convenience on which experimental designs capitalize. As shown in the case of cn and bw, all the alleles present in any chromosome in a male must be transmitted as a group, without being recombined with alleles present in the homologous chromosome.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The complete absence of crossing over in Drosophila males is atypical, but it is not unusual for frequencies of recombination to differ between the sexes. For example, the human genome also exhibits more recombination in females than in males, but the difference is not nearly so extreme as that in Drosophila. On the average, across the entire human genome, the recombination frequency between genetic markers in males is approximately 60 percent of the value observed between the same genetic markers in females.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

SUMMING UP ■ Gametes produced as a result of meiosis can be either parental or recombinant. ■ If recombination between two genes occurs less than 50 percent of the time, then the genes are linked. ■ Recombinant gametes of linked genes are created as a result of crossing over in meiosis. ■ Each pair of loci has a characteristic recombination frequency, and it does not matter whether the alleles of those loci are in cis or trans genotypes in the parental generations. ■ If two genes are syntenic but undergo recombination more than 50 percent of the time, then they assort independently, just as though they were on different chromosomes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Recombination rates can differ between males and females

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

5.2 Genetic Mapping The linkage of the genes in a chromosome can be represented in the form of a genetic map, which shows the linear order of the genes along the chromosome with the distances between adjacent genes proportional to the frequency of recombination between them. A genetic map is also called a linkage map or a chromosome map. The concept of genetic mapping was first

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

developed by Morgan's student Alfred H. Sturtevant in 1913 (see Roots of Discovery: Genes All in a Row). The early geneticists understood that recombination between genes takes place by an exchange of segments between homologous chromosomes in the process now called crossing over. Each crossover is manifested physically as a chiasma, or crossshaped configuration, between homologous chromosomes; chiasmata are observed in prophase I of meiosis. Each chiasma results from the breaking and rejoining of nonsister chromatids, with the result that an exchange of corresponding segments takes place between them. The theory of crossing over is that each chiasma results in a new association of genetic markers. This process is illustrated in FIGURE 5.3. When there is no crossing over (Figure 5.3A), the alleles present in each homologous chromosome remain in the same combination. When crossing over does take place (Figure 5.3B), the outermost alleles in two of the chromatids are interchanged (recombined).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.



Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.3 (A) Alleles in the same chromosome remain together when there is no crossing over between them. (B) When there is crossing over, the result is two recombinant and two nonrecombinant products, because the exchange is between only two of the four chromatids.

Map Distance and Frequency of Recombination The unit of distance in a genetic map is called a map unit. Over short intervals, 1 map unit equals 1 percent recombination. For example, two genes that recombine with a frequency of 3.5 percent are said to be located 3.5 map units apart. One map unit is

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

also called a centimorgan (abbreviated cM) in honor of T. H. Morgan. A distance of 3.5 map units therefore equals 3.5 centimorgans and indicates 3.5 percent recombination between the genes. For ease of reference, we list the four completely equivalent ways in which a genetic distance between two genes may be represented: ■ As a frequency of recombination (in the foregoing example, 0.035) ■ As a percent recombination (3.5 percent) ■ As a map distance in map units (3.5 map units) ■ As a map distance in centimorgans (3.5 cM)

ROOTS OF DISCOVERY Genes All in a Row Alfred H. Sturtevant (1913) Columbia University, New York, New York

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The Linear Arrangement of Six SexLinked Factors in Drosophila, as Shown by Their Mode of Association Genetic mapping remains the cornerstone of genetic analysis. It is the principal technique used in modern human genetics to identify the chromosomal location of mutant genes associated with inherited diseases. The genetic markers used in human genetics are often single-nucleotide polymorphisms (SNPs) that differ from one person to the next, but the basic principles of genetic mapping are the same as those originally enunciated by Sturtevant. In this excerpt, we have substituted the symbols

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

presently in use for the genes, y (yellow body), w (white eyes), v (vermilion eyes), m (miniature wings), and r (rudimentary wings). (The sixth gene mentioned is another mutant allele of white, now called white-eosin.) In this paper, Sturtevant uses the term “crossing over” instead of “recombination,” and “crossovers” instead of “recombinant chromosomes.” We have retained his original terms but, in a few cases, have put the modern equivalent in brackets.

“These results form a new argument in favor of the chromosome view of inheritance, since they strongly indicate

”

that the factors investigated are arranged in a linear series.

After summarizing Morgan's observations with respect to w and m, Sturtevant noted: It became evident that some of the sex-linked factors are associated, i.e., that crossing over does not occur freely between some factors, as shown by the fact that the combinations present in the F1 flies are much more frequent in

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the F2 than are new combinations of the same characters. This means, on the chromosome view, that the chromosomes, or at least certain segments of them, are much more likely to remain intact during meiosis than they are to interchange materials…. It would seem, if this hypothesis be correct, that the proportion of “crossovers” [recombinant chromosomes] could be used as an index of the distance between any two factors. Then, by determining the distances (in the above sense) between A and B and between B and C, one should be able to predict AC…. Is this prediction borne out? In the data in the following table, the first column is the recombination frequency predicted based on the sum of the frequency of recombination between intervening markers (that is, the sum of the observed

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

frequencies between markers A and B and between markers B and C). The second column is the observed frequency of recombination between A and C, without taking B into account. Factors

Calculated distance

Observed percentage of crossovers

y–v

30.7

32.2

y–m

33.7

35.5

y–r

57.6

37.6

w–m

32.7

33.7

w–r

56.6

45.2

Reproduced from Sturtevant, Z. H. (1913). The linear arrangement of six sex-linked factors in Drosophila, as shown by their mode of association. Journal of Experimental Zoology, 14(1), 43–59. Copyright (c) 2005 by John Wiley and Sons, Inc. Reproduced with permission of John Wiley & Sons, Inc.

Sturtevant notes:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

It will be noticed at once that the longer distances, y–r and w–r, give a smaller percent of crossovers than the calculation calls for. This is a point which was to be expected and is probably due to the occurrence of two breaks in the same chromosome, or “double crossing over.” The final conclusion of the paper is its most significant, in that it states the conceptual basis for mapping genes based on recombination frequencies that remains the paradigm today: It has been found possible to arrange six sex-linked factors in Drosophila in a linear series, using the number of crossovers per 100 cases [the frequency of recombination] as an index of the distance between any two factors. A source of error in predicting the strength of association between untried factors is found in

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

double crossing over. The occurrence of this phenomenon is demonstrated…. These results form a new argument in favor of the chromosome view of inheritance, since they strongly indicate that the factors investigated are arranged in a linear series. Source: A. H. Sturtevant, J. Exp. Zool. 14 (2005): 43–59.

Physically, 1 map unit can be defined as the length of the chromosome in which, on average, 1 crossover is formed in every 50 cells undergoing meiosis. This principle is illustrated in FIGURE 5.4. If 1 meiotic cell in 50 has a crossover, the frequency of crossing over equals 1/50, or 2 percent. The frequency of recombination between the genes, however, is 1 percent. The correspondence of 1 percent recombination with 2 percent crossing over is a little confusing until you realize that each crossover results in 2 recombinant chromatids and 2 nonrecombinant chromatids (Figure 5.4). A frequency of crossing over of 2 percent means that, among the 200 chromosomes that result from meiosis in 50 cells, exactly 2 chromosomes (those involved in the exchange) are recombinant for genetic markers flanking the crossover. In other words, 2 percent crossing over corresponds to 1 percent recombination because only half the chromatids in each cell with

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

an exchange are actually recombinant.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.4 Chromosomal configurations in 50 meiotic cells, in which 1 cell has a crossover between two genes. (A) The 49 cells without a crossover result in 98 A B and 98 a b chromosomes; these are all nonrecombinant. (B) The cell with a crossover yields chromosomes that are A B, A b, a B, and a b, of which the middle two types are recombinant chromosomes. (C) The recombination frequency equals 2/200, or 1 percent (also called 1 map unit or 1 cM). Therefore, 1 percent recombination means that 1 meiotic cell in 50 has a crossover in the region between the genes.

In situations in which there are genetic markers along the chromosome, such as the A, a and B, b pairs of alleles in Figure 5.4, recombination between the marker genes takes place only when crossing over occurs between the genes. FIGURE 5.5 illustrates a case in which a crossover takes place between the gene A and the centromere, rather than between the genes A and B. The crossover does result in the physical exchange of segments between the innermost chromatids. Because the exchange is located outside the region between A and B, however, all of the resulting gametes must carry either the A B or the a b allele

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

combination. These are nonrecombinant chromosomes. The presence of the crossover is undetected because it does not occur in the region between the genetic markers.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.5 Crossing over outside of the region between two genes is not detectable through recombination. Although a segment of chromosome is exchanged, the genetic markers stay in the nonrecombinant configurations—in this case, A B and a b.

In some cases, the region between genetic markers is large enough that two (or even more) crossovers can be formed in a single meiotic cell. One possible configuration for two crossovers is shown in FIGURE 5.6. In this example, both crossovers are between the same pair of chromatids. As a result, a physical exchange of a segment of chromosome occurs between the marker genes, but the double crossover remains undetected because the markers themselves are not recombined. The absence of recombination results from the fact that insofar as recombination between A and B is concerned, the second crossover reverses the effect of the first. The resulting chromosomes are either A B or a b, both of which are nonrecombinant.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

FIGURE 5.6 If two crossovers take place between marker genes, and both involve the same pair of chromatids, then neither crossover is detected because all of the resulting chromosomes are nonrecombinant A B or a b.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Given that double crossing over in a region between two genes can remain undetected because it does not result in recombinant chromosomes, there is an important distinction between the distance between two genes as measured by the recombination frequency and that distance as measured in map units. Map units measure how much crossing over takes place between the genes. For any two genes, the map distance between them is defined as follows:

In contrast, the recombination frequency reflects how much recombination is actually observed in a particular experiment. When double crossovers do not yield recombinant gametes, such as the one in Figure 5.6, they contribute to the map distance but do not contribute to the recombination frequency. This distinction is important only when the region in question is large enough that double crossing over can occur. If the region between the genes is so short that no more than one crossover can be formed in the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

region in any one meiosis, then map units and recombination frequencies are the same (because there are no multiple crossovers that can undo each other). This is the basis for saying that, for short distances, 1 map unit equals 1 percent recombination. Over an interval so short as to yield 1 percent observed recombination, the frequency of multiple crossovers is usually negligible, so the map distance equals the recombination frequency in this case.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

For a series of linked genes, if the chromosomal region between each adjacent pair is sufficiently short that multiple crossovers are not formed in the region, the recombination frequencies (and hence the map distances) between the genes are additive. This important feature of recombination, which is also the logic used in genetic mapping, is illustrated in FIGURE 5.7. The genes are all in the X chromosome of Drosophila: y (yellow body), rb (ruby eye color), and cv (shortened wing cross-vein). The recombination frequency between genes y and rb is 7.5 percent, and that between rb and cv is 6.2 percent. The genetic map might be any one of three possibilities, depending on whether y, cv, or rb is in the middle. Map A, which has y in the middle, can be excluded because it implies that the recombination frequency between rb and cv should be greater than that between rb and y, which contradicts the observed data.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.7 In Drosophila, the genes y (yellow body) and rb (ruby eyes) have a recombination frequency of 7.5 percent, and rb and cv (shortened wing cross-vein) have a recombination frequency of 6.2 percent. There are three possible genetic maps, depending on whether y is in the middle (A), cv is in the middle (B), or rb is in the middle (C). Map A can be excluded because it implies that rb and y are closer than rb and cv, whereas the observed recombination frequency between rb and y is larger than that between rb and cv. Maps B and C are compatible with the data given.

Maps B and C are both consistent with the recombination frequencies. They differ in their predictions regarding the recombination frequency between y and cv. In map B, the predicted distance is 1.3 map units; by comparison, in map C, the predicted distance is 13.3 map units. In reality, the observed recombination frequency between y and cv is 13.3 percent. Map C is therefore correct. There are actually two genetic maps corresponding to map C. They differ only in whether y is placed at the left or at the right. One map is:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

The other map is:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

These two ways of depicting the genetic map are completely equivalent. By this type of reasoning, all the known genes in a chromosome can be assigned a position in a genetic map of the chromosome by considering successive subsets of three markers. Each set of syntenic genes constitutes a linkage group. The number of linkage groups is the same as the haploid number of chromosomes of the species. For example, cultivated corn (Zea mays) has 10 pairs of chromosomes and 10 linkage groups. A partial genetic map of chromosome 10 is shown in FIGURE 5.8, along with the phenotypes shown by some of the mutants. The ears of corn in Figures 5.8C and 5.8F demonstrate the result of Mendelian segregation. Figure 5.8C shows a 3 : 1 segregation of yellow : orange kernels produced by the recessive orange pericarp-2 (orp2) allele in a cross between two heterozygous genotypes, and Figure 5.8F shows a 1 : 1 segregation of marbled : white kernels produced by the dominant allele R1-mb in a cross between a heterozygous genotype and a homozygous normal genotype. As an exercise, try counting the kernels in Figures 5.8C and 5.8F to convince yourselves that these ratios hold.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.8 Genetic map of chromosome 10 of corn, Zea mays. The map distance to each gene is given in standard map units (centimorgans) relative to a position 0 for the telomere of the short arm (lower left). (A) Mutations in the gene oil yellow-1 (oy1) result in a yellow-green plant. The plant in the foreground is heterozygous for the dominant allele Oy1; behind is a normal plant. (B) Mutations in the gene lesion-16 (les16) result in many small to medium-sized,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

irregularly spaced, discolored spots on the leaf blade and sheath. The photograph shows the phenotype of a heterozygote for Les16, a dominant allele. (C) The orp2 allele is a recessive allele expressed as orange pericarp, a maternal tissue that surrounds the kernels. The ear shows the segregation of orp2 in a cross between two heterozygous genotypes, yielding a 3 : 1 ratio of yellow : orange seeds. (D) The gene zn1 is zebra necrotic-1, in which dying tissue appears in longitudinal leaf bands. The leaf on the left is homozygous zn1; that on the right is wild type. (E) Mutations in the gene teopod-2 (tp2) result in many small, partially podded ears and a simple tassel. An ear from a plant heterozygous for the dominant allele Tp2 is shown. (F) The mutation R1-mb is an allele of the r1 gene, resulting in red or purple color in the aleurone layer of the seed. Note the marbled color in kernels of an ear segregating for R1-mb. Data from an illustration by E. H. Coe. Photographs courtesy of M. G. Neuffer, College of

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Agriculture, Food, and Natural Resources, University of Missouri.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Crossing Over The orderly arrangement of genes represented by a genetic map is consistent with the conclusion that each gene occupies a welldefined site (locus) in the chromosome. In a heterozygote, the alleles occupy corresponding locations in the homologous chromosomes. Crossing over is brought about by a physical

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

exchange of segments resulting in a new association of genes in the same chromosome. This process has the following features: 1. The exchange of segments between parental chromatids takes place in the first meiotic prophase, after the chromosomes have duplicated. At this stage, the four chromatids (strands) of a pair of homologous chromosomes are closely synapsed. Crossing over is a physical exchange between chromatids in a pair of homologous chromosomes. 2. The exchange process consists of the breaking and rejoining of the two chromatids, resulting in the reciprocal exchange of equal and corresponding segments between them (Figure 5.3). 3. The sites of crossing over are more or less random along the length of a chromosome pair. Hence, the probability of crossing over between two genes increases as the physical distance between the genes along the chromosome becomes larger. This principle is the basis of genetic mapping.

Recombination Between Genes Results from a Physical Exchange Between Chromosomes A classic demonstration that recombination is associated with a physical exchange of segments between homologous

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

chromosomes made use of two structurally altered X chromosomes of Drosophila that permitted parental and recombinant chromosomes to be recognized in a microscope. This experiment, performed in 1936 by Curt Stern, is outlined in FIGURE 5.9. One of the abnormal X chromosomes was missing a segment that had become attached to chromosome 4; this X chromosome could be identified by its missing terminal segment. The second aberrant X chromosome could be identified because it had a small second arm consisting of a piece of the Y

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

chromosome. The mutant alleles car (a recessive allele resulting in carnation eye color instead of wild-type red) and B (a dominant allele resulting in bar-shaped eyes instead of round) were present in the first abnormal X chromosome, and the wild-type alleles of these genes were in the second abnormal X.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.9 (A) Diagram of a cross in which the two X chromosomes in a Drosophila female are morphologically distinguishable from each other and from a normal X chromosome. One X chromosome has a missing terminal segment, and the other has a second arm consisting of a fragment of the Y chromosome. (B) Result of the cross. The offspring that are carnation but not bar (single asterisks) contain a structurally normal X chromosome, and the offspring that are bar but not carnation (double asterisks) contain an X chromosome with both morphological markers. The result demonstrates that genetic recombination between marker genes is associated with physical exchange between homologous chromosomes. Segregation of the missing terminal segment of the X chromosome, which is attached to chromosome 4, is not shown.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Females with the two structurally and genetically marked X chromosomes were mated with males having a normal X that carried the recessive alleles of the genes (Figure 5.9). Among the progeny from this cross, flies with parental or recombinant chromosomes could be recognized by their eye color and shape. In the genetically recombinant progeny from the cross, the X chromosome had the appearance that would be expected if recombination of the genes were accompanied by an exchange that recombined the chromosome markers. In particular, the progeny with wild-type, bar-shaped eyes had an X chromosome with a missing terminal segment and the attached Y arm; in contrast, the progeny with carnation-colored, round eyes had a structurally normal X chromosome with no missing terminus and no Y arm. As expected, the nonrecombinant progeny were found to have an X chromosome structurally identical to one of the X chromosomes in the mother.

Crossing Over Takes Place at the Four-

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Strand Stage of Meiosis So far we have asserted, without citing experimental evidence, that crossing over takes place in meiosis after the chromosomes have duplicated, at the stage when each bivalent has four chromatid strands. One experimental proof that crossing over takes place after the chromosomes have duplicated came from a study of laboratory stocks of D. melanogaster in which the two X chromosomes in a female are joined to a common centromere to form an aberrant chromosome called an attached-X, or compound-X, chromosome. The normal X chromosome in Drosophila has a centromere almost at the end of the chromosome. The attachment of two of these chromosomes to a single centromere results in a chromosome with two equal arms, each consisting of a virtually complete X. Females with a

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

compound-X chromosome usually contain a Y chromosome as well, and they produce two classes of viable offspring: females that have the maternal compound-X chromosome along with a paternal Y chromosome, and males that have the maternal Y chromosome along with a paternal X chromosome (FIGURE 5.10).

FIGURE 5.10 Attached-X (compound-X) chromosomes in Drosophila. (A) A structurally normal X chromosome in a female. (B) An attached-X chromosome, with the long arms of two normal X chromosomes attached to a common centromere. (C) Typical attached-X females also contain a Y chromosome. (D) Outcome of a cross between an attached-X female and a normal male. Genotypes with either three X chromosomes or no X chromosomes are lethal. In this case, a male fly receives its X chromosome from its father and its Y chromosome from its mother—the opposite of the usual situation in Drosophila.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Attached-X chromosomes are frequently used to study X-linked genes in Drosophila. When an attached-X female is mated with a male carrying any X-linked mutation, the sons receive the mutant X chromosome and the daughters receive the attached-X chromosome. In matings with attached-X females, therefore, the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

inheritance of an X-linked gene in the male passes from father to son to grandson, and so forth, which is the opposite of usual Xlinked inheritance. In an attached-X chromosome in which one X carries a recessive allele and the other carries the wild-type nonmutant allele, crossing over between the X-chromosome arms can yield attached-X products in which the recessive allele is present in both arms of the attached-X chromosome (FIGURE 5.11). Hence, attached-X females that are heterozygous can produce some female progeny that are homozygous for the recessive allele. The frequency with which homozygosity is observed increases with increasing map distance of the gene from the centromere. From the diagrams in Figure 5.11, it is clear that homozygosity can result only if the crossover between the gene and the centromere takes place after the chromosome has duplicated. The finding of homozygous attached-X female progeny therefore implies that crossing over takes place at the four-strand stage of meiosis. If this were not the case, and crossing over happened before duplication of the chromosome (at the two-strand stage), it would result in a swap of the alleles between the chromosome arms and would never yield the homozygous products that are actually observed.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

FIGURE 5.11 Crossing over must take place at the four-strand stage in meiosis to produce a homozygous attached-X chromosome from one that is heterozygous for an allele. To yield homozygosity, the exchange must take place between the centromere and the gene.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Multiple Crossovers When two genes are located far apart along a chromosome, more than one crossover can occur between them in a single meiosis, and the probability of multiple crossovers increases with the distance between the genes. Multiple crossing over complicates the interpretation of recombination data. The problem is that some of the multiple exchanges between genes do not result in recombination and, therefore, are not detected. As we saw in Figure 5.6, the effect of one crossover can be canceled by another

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

crossover further along the chromosome. If two exchanges between the same two chromatids take place between the genes A and B, then their net effect will be that all chromosomes are nonrecombinant, either A B or a b. Two of the products of this meiosis have an interchange of their middle segments, but the chromosomes are not recombinant for the genetic markers, so they are genetically indistinguishable from noncrossover chromosomes. The possibility of such canceling means that the true map distance, defined in Equation 1, is underestimated by setting it equal to the observed frequency of recombination. In higher organisms, double crossing over is effectively precluded in chromosome segments that are sufficiently short. Therefore, by using recombination data for closely linked markers to build up genetic linkage maps, we can avoid multiple crossovers that cancel each other's effects. For genes that are distant along a chromosome, the true map distance is estimated by summing the recombination frequencies across shorter subintervals within which multiple crossovers do not occur.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The minimum recombination frequency between two genes is 0. The recombination frequency also has a maximum: No matter how far apart two genes may be, the maximum frequency of recombination between any two genes is 50 percent. This maximum recombination frequency is presumably what pertained with respect to the syntenic genes Mendel analyzed (Table 5.1); 50 percent recombination is the same value that would be observed if the genes were on nonhomologous chromosomes and assorted independently.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

The maximum frequency of recombination is observed when the genes are so far apart in the chromosome that at least one crossover nearly always forms between them. Figure 5.3B shows that a single exchange in every meiosis would result in half of the products having parental combinations and the other half having recombinant combinations of the genes. Two exchanges between two genes have the same effect, as shown in FIGURE 5.12. Figure 5.12A shows a two-strand double crossover, in which the same chromatids participate in both exchanges; no recombination

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

of the marker genes is detectable. When the two exchanges have one chromatid in common (three-strand double crossover, Figures 5.12B and C), the result is indistinguishable from that of a single exchange: Two products with parental combinations and two with recombinant combinations are produced. Note that there are two types of three-strand doubles, depending on which three chromatids participate. The final possibility is that the second exchange involves the chromatids that did not participate in the first exchange (four-strand double crossover, Figure 5.12D), in which case all four products are recombinant.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.





Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.



FIGURE 5.12 The result of two crossovers in the interval between two genes is indistinguishable from independent assortment of the genes, provided that the chromatids participate at random in the exchanges. (A) A two-strand double crossover. (B, C) The two types of three-strand double crossovers. (D) A four-strand double crossover.

In most organisms, when double crossovers form, the chromatids that take part in the two exchange events are selected at random. In this case, the expected proportions of the three types of double exchanges are 1/4 two-strand doubles, 1/2 three-strand doubles, and 1/4 four-strand doubles. This means that, on average,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

recombinant chromatids will be found among the four chromatids

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

produced from meioses with two exchanges between a pair of genes. This is the same proportion obtained with a single exchange between the genes. Moreover, a maximum of 50 percent recombination is obtained for any number of exchanges. In the discussion of Figure 5.12, we emphasized that in most organisms, the chromatids taking part in double-exchange events are selected at random. In such a case, the maximum frequency of recombination is 50 percent. When there is a nonrandom choice of chromatids in successive crossovers, the phenomenon is called chromatid interference. In Figure 5.12, notice that, relative to a random choice of chromatids, an excess of four-strand double crossing over (positive chromatid interference) results in a maximum frequency of recombination greater than 50 percent; likewise, an excess of two-strand double crossing over (negative chromatid interference) results in a maximum frequency of recombination less than 50 percent. Therefore, the finding that the maximum frequency of recombination between two genes in the same chromosome is not 50 percent can be regarded as evidence for chromatid interference. Positive chromatid interference has not yet been observed in any organism; negative chromatid interference has been reported in some fungi. Double crossing over is detectable in recombination experiments that employ three-point crosses, which include three pairs of alleles. If a third pair of alleles, c+ and c, is located between the two with which we have been concerned (the outermost genetic markers), then double exchanges in the region can be detected when the crossovers flank the c gene (FIGURE 5.13). The two crossovers—which in this example take place between the same +

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

pair of chromatids—would result in a reciprocal exchange of the c+ and c alleles between the chromatids. A three-point cross is an efficient way to obtain recombination data; it is also a simple method for determining the order of the three genes, as we shall see in the next section.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.13 A two-strand double crossover that spans the middle pair of alleles in a triple heterozygote results in a reciprocal exchange of the middle pair of alleles between the two participating chromatids.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

SUMMING UP ■ The frequency of crossing over between particular pairs of genes is proportional to the physical distance between them on the chromosome. ■ One map unit, or one centimorgan, corresponds to the distance for which crossing over produces 1 percent recombinant gametes. ■ A single crossover occurring within a meiosis produces two nonrecombinant and two recombinant gametes. ■ Two or more crossovers can occur between pairs of loci.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Crossing over occurs after DNA replication, during prophase I of meiosis.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

5.3 Genetic Mapping in a Three-Point Testcross The data in TABLE 5.3 result from a testcross in corn with three linked genes. We will use these data to illustrate the analysis of a three-point cross. The recessive alleles are lz (for lazy or prostrate growth habit), gl (for glossy leaf), and su (for sugary endosperm), and the multiply heterozygous parent in the cross had the following genotype:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

TABLE 5.3 Progeny from a three-point testcross in corn Phenotype of testcross

Genotype of gamete from hybrid

Number

progeny

parent

Normal (wild type)

Lz Gl Su

286

Lazy

lz Gl Su

33

Glossy

Lz gl Su

59

Sugary

Lz Gl su

4

Lazy, glossy

lz gl Su

2

Lazy, sugary

lz Gl su

44

Glossy, sugary

Lz gl su

40

Lazy, glossy, sugary

lz gl su

272

Thus, the two classes of progeny that inherit noncrossover (parental-type) gametes are the normal plants and plants with the lazy-glossy-sugary phenotype. These classes are far larger than any of the crossover classes. If the combination of dominant and

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

recessive alleles in the chromosomes of the heterozygous parent was unknown, then we could deduce from their relative frequencies among the progeny that the noncrossover gametes were Lz Gl Su and lz gl su. This point is important enough to state as a general principle: In any genetic cross involving linked genes, no matter how complex, the two most frequent types of gametes with respect to any pair of genes are

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

nonrecombinant; these provide the linkage phase (cis versus trans) of the alleles of the genes in the multiply heterozygous parent. In mapping experiments, the order of genes along the chromosome is usually not known in advance. In Table 5.3, the order in which the three genes are shown is entirely arbitrary. There is, however, an easy way to determine the correct order from three-point data: Simply identify the genotypes of the double-crossover gametes produced by the heterozygous parent and compare them with the nonrecombinant gametes. Because the probability of two simultaneous exchanges is considerably smaller than the probability of either single exchange, the double-crossover gametes will be the least frequent types. Table 5.3 shows that the classes composed of four plants with the sugary phenotype and two plants with the lazy—glossy phenotype (products of the Lz Gl su and lz gl Su gametes, respectively) are the least frequent and, therefore, constitute the double-crossover progeny. The effect of double crossing over, as Figure 5.13 shows, is to interchange the members of the middle pair of alleles between the chromosomes. Thus, if the parental chromosomes are:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

and the double-crossover chromosomes are:

then Su and su are interchanged by the double crossing over and must be the middle pair of alleles. Therefore, the genotype of the heterozygous parent in the cross should be written as:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

which is now diagrammed correctly with respect to both the order of the genes and the configuration of alleles in the homologous chromosomes. A two-strand double crossover between chromatids of these parental types is diagrammed next, and the products can be seen to correspond to the two types of gametes identified in the data as the double crossovers.

From this diagram, it can also be seen that the reciprocal products of a single crossover between lz and su would be Lz su gl and lz Su Gl, and that the products of a single exchange between su and gl would be Lz Su gl and lz su Gl. We can now summarize the data in a more informative way, writing the genes in correct order and identifying the numbers of the different chromosome types produced by the heterozygous parent that are present in the progeny.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Note that each class of single recombinants includes two reciprocal products, and that these are found in approximately equal frequencies (40 versus 33, and 59 versus 44). This observation illustrates an important principle: The two reciprocal products that result from any crossover, or any combination of crossovers, are expected to appear in approximately equal frequencies among the progeny.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In calculating the frequency of recombination from the data, remember that the double-recombinant chromosomes result from two exchanges—one in each of the chromosome regions defined by the three genes. Therefore, chromosomes that are recombinant between lz and su are represented by the following chromosome types:

Because 79/740, or 10.7 percent, of the chromosomes recovered in the progeny are recombinant between the lz and su genes, the map distance between these genes is estimated as 10.7 map units, or 10.7 centimorgans.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Similarly, the chromosomes that are recombinant between su and gl are represented by the following chromosome types:

The recombination frequency between the su and gl genes is 109/740, or 14.7 percent, so the map distance between them is estimated as 14.7 map units, or 14.7 centimorgans. The genetic map of the chromosome segment in which the three genes are located is therefore:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

One error that many students make in analyzing three-point crosses is to forget to include the double recombinants when calculating the recombination frequency between adjacent genes. You can keep from falling into this trap by remembering that the double-recombinant chromosomes have single recombination in both regions.

Chromosome Interference in Double Crossovers The ability to detect double crossovers makes it possible to determine whether exchanges in two different regions occur independently. Using the information from the example with corn, we know from the recombination frequencies that the probability of recombination is 0.107 between lz and su and 0.147 between su and gl. If crossing over is independent in the two regions (which

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

means that the formation of one exchange does not alter the probability of the second exchange), then the probability of an exchange in both regions is the product of these separate probabilities, or 0.107 × 0.147 = 0.0157. This implies that in a sample of 740 gametes, the expected number of double crossovers would be 740 × 0.0157 = 11.6, whereas the number actually observed is only 6. Such deficiencies in the observed number of double crossovers are common and identify a phenomenon called chromosome interference, in which crossing over in one region of a chromosome reduces the probability of a second crossover in a nearby region. Because chromosome interference is nearly universal, whereas chromatid interference is virtually unknown, the term interference, when used without qualification, nearly always refers to chromosome interference.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The coefficient of coincidence is the observed number of doublerecombinant chromosomes divided by the expected number. Its value provides a quantitative measure of the degree of interference, defined as:

For the data in our example, the coefficient of coincidence is 6/11.6 = 0.51, which means that the observed number of double crossovers was only 51 percent of the number we would expect to observe if crossing over in the two regions were independent. The degree of interference depends on the distance between the genetic markers and on the species. In some species, interference increases as the distance between the two outside markers becomes smaller, until a point is reached at which double crossing over is eliminated—that is, no double crossovers are found, and the coefficient of coincidence equals 0 (or, to say the same thing, the interference equals 1). In Drosophila, this distance is approximately

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

10 map units. In most organisms, when the total distance between the genetic markers is greater than approximately 30 map units, interference essentially disappears and the coefficient of coincidence approaches 1.

Genetic Mapping Functions The effect of interference on the relationship between genetic map distance and the frequency of recombination is illustrated in

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.14. Each curve is an example of a mapping function, which is the mathematical relationship between the genetic distance across an interval in map units (centimorgans) and the observed frequency of recombination across the interval. In other words, a mapping function converts a map distance between genetic markers into a recombination frequency between the markers. As we have seen, when the map distance between the markers is small, the recombination frequency equals the map distance. This principle is reflected in the curves in Figure 5.14 in the region in which the map distance is smaller than approximately 15 cM. At less than this distance, all of the curves are nearly straight lines, which means that map distance and recombination frequency are equal: 1 map unit equals 1 percent recombination, and 10 map units equals 10 percent recombination.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.14 A mapping function is the relation between genetic map distance across an interval and the observed frequency of recombination across the interval. Map distance is defined as onehalf the average number of crossovers converted into a percentage. The three mapping functions correspond to different assumptions about interference, i. In the first curve, i = 1 (complete interference); in this case, the curve terminates abruptly because there can be at most one crossover per chromosome. In the third curve, i = 0 (no interference). The mapping function in the middle is based on the assumption that i decreases as a linear function of distance.

For distances greater than 15 map units, the curves differ. The upper curve is based on the assumption of complete interference i, so that i = 1. With this mapping function, the linear relation holds all the way to a map distance of 50 cM. The curve then terminates abruptly at a map distance of 50 map units. It terminates because complete interference means that a chromosome can have no more than one crossover along its length; because one crossover corresponds to 50 map units, this is the maximum length of the chromosome.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

The other two curves in Figure 5.14 are for different patterns of interference along the chromosome. The bottom curve is usually called Haldane's mapping function, after its inventor. It assumes no interference (i = 0), and the mathematical form of the function is r = (1/2) (1 − e−d/50), where d is the map distance in centimorgans. Any mapping function for which i is between 0 and 1 must lie in the interval between the top and bottom curves. The particular example shown in Figure 5.14 is Kosambi's mapping function, in which interference is assumed to decrease as a linear function of distance according to i = 1 − 2r. Although simple in its underlying assumptions, the formula for Kosambi's function is not simple.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Most mapping functions are almost linear near the origin, as are those in Figure 5.14. This near linearity implies that for map distances smaller than about 15 cM, whatever the pattern of chromosome interference, there are so few double recombinants that the recombination frequency in percent essentially equals the map distance. Hence, the map distance between two widely separated genetic markers can be estimated with some confidence by summing the map distances across smaller segments between the markers, provided that each of the smaller segments is less than about 15 map units in length.

Genetic Map Distance and Physical Distance Generally speaking, the greater the physical separation between genes along a chromosome, the greater the map distance between them. On the one hand, physical distance and genetic map distance are usually correlated, in that a greater distance between genetic markers affords a greater chance for crossing over to take place.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

On the other hand, the general correlation between physical distance and genetic map distance is by no means absolute. We have already noted that the frequency of recombination between genes may differ in males and females. An unequal frequency of recombination means that the sexes can have different map distances in their genetic maps, although the physical chromosomes of the two sexes are the same and the genes must have the same linear order. An extreme example is found in Drosophila males, in which no crossing over occurs. In Drosophila

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

males, the map distance between any pair of genes located in the same chromosome is 0; nevertheless, genes in different chromosomes do undergo independent assortment. The general correlation between physical distance and genetic map distance can break down even in a single chromosome. For example, crossing over is much less frequent in some regions of the chromosome than in other regions. The term heterochromatin refers to certain regions of the chromosome that have a dense, compact structure in interphase; these regions take up many of the standard dyes used to make chromosomes visible. The rest of the chromatin, which becomes visible only after chromosome condensation in mitosis or meiosis, is called euchromatin. In most organisms, the major heterochromatic regions are adjacent to the centromere; smaller blocks are present at the ends of the chromosome arms (the telomeres) and interspersed with the euchromatin. In general, crossing over is much less frequent in regions of heterochromatin than in regions of euchromatin. Because less crossing over occurs in heterochromatin, a given length of heterochromatin will appear much shorter in the genetic map than an equal length of euchromatin. In heterochromatic regions, therefore, the genetic map gives a distorted picture of the physical map. An example of such distortion appears in FIGURE

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

5.15, which compares the physical map and the genetic map of chromosome 2 in Drosophila. The physical map is depicted as the chromosome appears in metaphase of mitosis. Two genes near the tips and two genes near the euchromatin–heterochromatin junction are indicated in the genetic map. The map distances across the euchromatic arms are 54.5 and 49.5 map units, respectively, for a total euchromatic map distance of 104.0 map units. However, the heterochromatin, which constitutes approximately 25 percent of the entire chromosome, has a genetic length of only 3.0 map units. The

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

distorted length of the heterochromatin in the genetic map results from the reduced frequency of crossing over in the heterochromatin. In spite of the distortion of the genetic map across the heterochromatin, in the regions of euchromatin there is a good correlation between the physical distance between genes and their distance in map units in the genetic map.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.15 Chromosome 2 in Drosophila as it appears in metaphase of mitosis (physical map, top) and in the genetic map (bottom). Heterochromatin and euchromatin are in contrasting colors. The genes are net (net wing veins), pr (purple eye color), cn (cinnabar eye color), and sp (speck of wing pigment). The total map length is 54.5 + 49.5 + 3.0 = 107.0 map units. The heterochromatin accounts for 3.0/107.0 = 2.8 percent of the total map length but constitutes approximately 25 percent of the physical length of the metaphase chromosome.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

SUMMING UP ■ A three-point testcross can be used to determine both the order of three genes on a chromosome and the map distances that separate them. ■ In the progeny of such a cross, the two reciprocal classes' progeny with parental gametes represent the largest class, while the double recombinants represent the smallest. ■ Comparison of the progeny with parental gametes to the progeny with double recombinants allows us to determine the order of the genes on the chromosomes. ■ Chromosome interference occurs when the observed number of double crossovers differs from the expected number, assuming crossover events occur independently of one another. It is quantified by the coefficient of coincidence. ■ Genetic mapping functions estimate the actual genetic map distance between genes based on the observed recombination frequency.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Genetic maps are generally correlated with physical maps of chromosomes, although significant differences in relative distances among loci do occur.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

5.4 Mapping by Tetrad Analysis Up to this point, all of our genetic analyses have involved making inferences from progeny of crosses, which of course resulted from the occurrence of many meioses (two for each individual). We were able to make some inferences about meiotic events based on the process as it occurs in the case of X chromosomal abnormalities (Figures 5.9 and 5.11), but they remain limited and indirect.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Some species of fungi allow us to examine the results of individual meioses directly. They are haploid, so recessive mutations are expressed, and they produce very large numbers of progeny. Most important, each meiotic tetrad is contained in a sac-like structure called an ascus (plural: asci) and can be recovered as an intact group. Each product of meiosis is included in a separate reproductive cell called an ascospore, and all of the ascospores formed from one meiotic cell remain together in the ascus (FIGURE 5.16). The advantage of using these organisms to study recombination is the potential for analyzing all of the products from each meiotic division. In this section, we show how the analysis of meiotic products in asci (tetrad analysis) can be used to study genetic linkage.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.16 Formation of an ascus containing all of the four products of a single meiosis. Each product of meiosis forms a reproductive cell called an ascospore; these cells are held together in the ascus. Segregation of one chromosome pair is shown.

In these organisms, the life cycle is relatively short. In budding yeast, the diploid is induced to undergo meiosis by growth in a medium that is deficient in its nutritional composition. The resulting haploid meiotic products, which form the ascospores, germinate to yield the vegetative stage of the organism (FIGURE 5.17). In some species, each of the four products of meiosis subsequently

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

undergoes a mitotic division, with the result being that each member of the tetrad yields a pair of genetically identical ascospores. In most of the organisms, the meiotic products, or their derivatives, are not arranged in any particular order in the ascus. However, bread molds of the genus Neurospora and related species have the useful characteristic that the meiotic products are arranged in a definite order directly related to the planes of the meiotic divisions. We will briefly discuss unordered tetrads and then turn to ordered ones, which are particularly valuable in the use of

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

genetic inference to understand meiosis.

FIGURE 5.17 Life cycle of the budding yeast Saccharomyces cerevisiae. Mating type is determined by the alleles a and α. Both haploid and diploid cells normally multiply by mitosis (budding). Depletion of nutrients in the growth medium induces meiosis and sporulation of cells in the diploid state. Diploid nuclei are orange; haploid nuclei are yellow.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Unordered Tetrads In the tetrads, when two pairs of alleles are segregating, three patterns of segregation are possible. For example, in the cross

there are three types of tetrads:

AB

Referred to as a parental ditype (PD) tetrad. Only two genotypes are

AB

represented, and the alleles have the same combinations found in the parents.

ab ab Ab

Referred to as a nonparental ditype (NPD) tetrad. Only two genotypes are

Ab

represented, but the alleles have nonparental combinations.

aB aB AB

Referred to as a tetratype (TT) tetrad. All four possible genotypes are

Ab

present.

aB

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

ab

The formation of tetrads, each containing all four products of a single meiosis, provides experimental proof of two fundamental principles of transmission genetics. The first principle is that of Mendelian segregation. Observe that each type of tetrad has a ratio of 2 A : 2 a and 2 B : 2 b. These ratios imply that Mendelian segregation is not merely a statistical average of ratios across a large number of meiotic divisions. Rather: Mendelian segregation of alleles takes place in each individual meiotic division and results in exactly two daughter cells carrying one allele and exactly two daughter cells carrying the alternative allele.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

The second principle relates to crossing over. Observe that tetratype tetrads are composed of the allele configurations A B, A b, a B, and a b. In other words, two of the products are parental types and two are recombinant types. The equality of parental and recombinant types implies that: Crossing over takes place at the four-strand stage of meiosis and results in two chromosomes that are recombinant and two chromosomes that are nonrecombinant. Tetrads have played an important role in the genetics of fungi because linkage between genes is easily identified through the analysis of tetrads. The detection of linkage is based on the relative numbers of parental ditype (PD) and nonparental ditype (NPD) tetrads:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

When genes are unlinked, the parental-ditype tetrads and nonparental-ditype tetrads are expected in equal frequencies (PD = NPD). Linkage is indicated when nonparental-ditype tetrads appear with a much lower frequency than parental-ditype tetrads (NPD < PD). The reason for the equality PD = NPD for unlinked genes is shown in FIGURE 5.18 (part A) for two pairs of alleles located in different chromosomes. In the absence of crossing over between either gene and its centromere, the two chromosomal configurations are equally likely at metaphase I, so PD = NPD. When there is crossing over between either gene and its centromere (Figure 5.18B), a tetratype tetrad results, but this does not change the fact that PD = NPD.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 5.18 Types of unordered asci produced with two genes in different chromosomes. (A) In the absence of crossing over, random arrangement of chromosome pairs at metaphase I results in two different combinations of chromatids—one yielding PD tetrads and the other NPD tetrads. (B) When crossing over takes place between one gene and its centromere, the two chromosome arrangements yield TT tetrads. If both genes are closely linked to their centromeres (so that crossing over between each gene and its centromere is rare), few TT tetrads are produced.

In contrast, when genes are linked, parental ditypes are far more frequent than nonparental ditypes. To see why, assume that the genes are linked and consider the events required for the production of the three types of tetrads. FIGURE 5.19 shows that when no crossing over takes place between the genes, a PD tetrad

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:18:49.

is formed. Single crossing over between the genes results in a TT tetrad. The formation of a two-strand, three-strand, or four-strand double crossover results in a PD, TT, or NPD tetrad, respectively. With linked genes, meiotic cells with no crossovers will always outnumber those with four-strand double crossovers. This results in

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the inequality NPD 1.0, which implies that amino acid changes take place faster than with neutrality—a result expected if amino acid changes were favorable. Examples with Ka/Ks > 1.0 include the sex-determining gene Sry, some homeodomain proteins, vertebrate growth hormone, some smell and taste receptors in certain species of Drosophila, and histone proteins specialized for spermatogenesis. We will discuss additional examples in the context of human evolution later in this chapter. The estimation and interpretation of the Ka/Ks ratio are not as

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

straightforward as they might first appear. Some of the complications are as follows: ■ Most first codon positions across a coding sequence consist of a combination of nondegenerate and twofold degenerate sites, and most third codon positions are a combination of nondegenerate, twofold, and fourfold degenerate sites. The proper enumeration of nonsynonymous sites and synonymous sites, therefore, requires an averaging procedure. In practice, the enumeration is usually carried out by computer programs. ■ The process of mutation may be biased. For example, most organisms show a mutational bias toward transitions over transversions. Any such bias can change the value of the neutral expectation of Ka/Ks. ■ Over sufficiently long periods of evolutionary time, mutations can take place at the same site, with one nucleotide changing

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

to another and then later changing to still another or back again to the original. Proper estimates of Ka/Ks need to be corrected for such “multiple hits.” ■ A value of Ka/Ks sufficiently greater than 1.0 to be statistically significant requires strong selection among many amino acids spread across the molecule. Amino acid changes may, therefore, be driven by selection but fail to produce a Ka/Ks significantly greater than 1.0. ■ Ka/Ks provides information only about possible selection at the amino acid level; it is completely insensitive to changes in level of gene expression. Hence, a gene may be under strong positive selection for regulatory changes but not show an elevated value of Ka/Ks.

Origins of New Genes: Orthologs and

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Paralogs In the course of evolution, new genes usually come from preexisting genes, and new gene functions evolve from previous gene functions. The raw material for new genes comes from duplications of regions of the genome, which may include one or more genes (described in the Human Karyotypes and Chromosome Behavior chapter). Duplications take place relatively frequently. Analysis of the genomic sequences of a wide variety of eukaryotes suggests that a eukaryotic genome containing 30,000 genes may be expected to undergo roughly 60–600 duplications per million years. From an evolutionary standpoint, two types of duplications need to be distinguished. The first is typified by beta-globin in the gene tree in FIGURE 19.1. Each time a speciation event took place, which is represented by a branching of the tree, the beta-globin gene

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

became duplicated in the sense that each derived species has a copy of the beta-globin gene that existed in the parental species. Genes that are duplicated as an accompaniment to speciation and that retain the same function are known as orthologous genes. New gene functions can arise from duplications that take place in the genome of a single species. Duplications within a genome result in paralogous genes. An example of paralogous genes are the beta-globin gene and the alpha-globin gene. Although the genes

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

are distinct now, they are sufficiently similar in sequence to show without ambiguity that they originated from a single gene in a remote common ancestor. Hence, the beta-globin and alpha-globin genes are paralogs. (To take this discussion one step further, the beta genes in any two mammalian species are orthologs, as are the alpha genes. However, the beta gene in one species and the alpha gene in another are paralogs.) When a gene duplication has taken place, the paralogs are redundant and one is free to evolve along any path. Probably the most common event is that one of the paralogs undergoes either a mutation that destroys its function or a deletion that eliminates it. Occasionally, however, mutations take place that cause the functions of the paralogs to diverge. They may evolve different pH optima, for example, so that one gene product performs optimally in compartments of the cell that are relatively basic and the other in compartments that are relatively acidic. In another case, genetic rearrangements may fuse two unrelated genes and yield a new activity. The Molecular Genetics of the Cell Cycle and Cancer chapter provides examples of how translocations that fuse a transcription factor with an oncogene can lead to acute leukemia. These particular rearrangements are extremely deleterious, but the creation of chimeric genes affords an example of how new functions can be acquired.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Gene duplications also allow the paralogous copies to evolve more specialized functions. An example is shown in FIGURE 19.5. At the top is a gene in a multicellular eukaryote that has enhancers for expression in tissue types 1 (yellow) and 2 (green). The other panels show a gene duplication followed by sequential mutations that knock out either the type-2 enhancer elements (left) or the type-1 enhancer elements (right). The result is that the paralog on the left is expressed only in tissue type 1 and that on the right only in tissue type 2. Specialization of paralogs accompanying loss of

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

functional capabilities is known as subfunctionalization. It may often be advantageous because each of the specialized genes is free to evolve toward optimal function in its own domain of expression.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.5 Specialization of paralogous genes by subfunctionalization. In this example, the paralogs become specialized for tissue-specific expression.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

SUMMING UP ■ Sequences of DNA and protein can be used to make inferences about evolutionary history. ■ Gene trees, which trace the history of single genes, are not necessarily the same as the phylogenetic trees of the taxa involved. ■ Several methods are available for inferring gene trees; bootstrapping provides a means of assessing the reliability of gene trees. ■ The concept of a “molecular clock” is useful in dating times of divergence in phylogenies. ■ Rates of evolution of particular nucleotides are affected by their functional role, especially in coding regions.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Gene duplication is an important source of new genes and functions.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

19.2 Ancient DNA Up to this point, all of our evolutionary inferences have derived from analysis of modern sequences. Sequencing of DNA from ancient material (aDNA), once considered a novelty, is rapidly becoming a powerful tool for evolutionary genetic analysis. The first such sequence reported was a 279-bp fragment of mitochondrial DNA from the skin of a 140-year-old museum specimen of a quagga, an extinct species of zebra. The first ostensibly ancient human DNA sequence reported was from an Egyptian mummy, dating to approximately 500 BCE. It was a sequence of a cloned 3.4-kb fragment; in fact, it likely originated from a modern contaminant. Subsequently, we have learned that sequencing aDNA has multiple challenges: ■ The age of the aDNA must be less than approximately 1 million years, and it must have been maintained in conditions appropriate for preservation.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Most aDNA is degraded into fragments of between 30 and 80 nucleotides. As such, it is difficult to work with, and assembling such fragments into longer contigs is challenging. ■ The bases in aDNA are degraded, and that degradation can affect the accuracy of sequencing. In particular, the deamination of cytosine to uracil (a topic examined in the Mutation, Repair, and Recombination chapter) means that, in the polymerase chain reaction (PCR) method used to amplify aDNA fragments, C-to-T and G-to-A transitions will occur, introducing errors (beyond those that occur in any PCR reaction) into the sequences determined. ■ Much of the DNA in typical ancient human samples is of microbial origin. Typically, in an ancient human sample, less than 20 percent of the DNA present is endogenous human DNA; the rest is primarily microbial.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

■ Contamination of samples is a critical problem, especially when working with human DNA. Since most aDNA methods involve PCR amplification as a first step, any contaminating modern DNA (perhaps from a skin cell of the investigator) will be an undamaged target for amplification. In fact, a number of methods have been developed to address these problems, some of which actually exploit the nature of the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

problem itself. In this section, we will outline a series of steps that an investigator might follow to obtain high-quality sequence information from ancient samples. Depending on the nature of the material and the questions being addressed, the goal might range from a set of single-nucleotide polymorphism (SNP) markers sufficient for phylogenetic reconstruction to a draft genome sequence that could be used to address multiple biological questions. As shown in FIGURE 19.6, ancient DNA typically has singlestranded ends, and it is in those ends where the uracil bases resulting from deamination of cytosine are concentrated. The single-stranded ends can be made double stranded with DNA polymerase, and the uracil bases removed by treatment with uracil N-glycosylase (with DNA polymerase filling in the resulting gap). Double-stranded oligonucleotide adapters are then ligated onto the ends of the fragments (FIGURE 19.6), which can subsequently be used as primer sequences for PCR amplification, creating an amplified library.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.6 Amplification of ancient DNA. (A) aDNA typically is found as double-stranded molecules with lengthy singlestranded ends, containing numerous deaminated cytosines. (B) Those ends are treated with uracil-N-glycosylase and are then filled in with DNA polymerase. (C) Double-stranded oligonucleotide linkers are added to the ends of the molecules to serve as priming sites for PCR. Reproduced from L. Orlando, T. P. Gilbert and E. Willersley, 2015, Nature Reviews:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Genetics, 16, pp. 395–405.

At this point, with massively parallel sequencing, it is technically possible to “shotgun” sequence the entire amplified library and then use bioinformatics tools to identify endogenous sequences. More commonly (and especially if the goal is not to assemble a complete genomic sequence), the library is enriched by a procedure such as that shown in FIGURE 19.7. In essence, this process involves selection of those sequences from an amplified aDNA library that hybridize with RNA sequences generated through in vitro transcription of modern human DNA sequences. The hybridized RNA can then be digested with RNase, leaving a pool of enriched single-stranded DNA that can be used as a template for massively parallel sequencing.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.7 Enrichment of endogenous sequences in aDNA. (A) Amplified aDNA, containing both endogenous and exogenous sequences, is denatured. (B) An amplified library obtained from modern DNA is prepared using phage T7 promoters as terminal adapters. That amplified DNA is used as a template for transcription of complementary RNA in the presence of biotinylated uracil. (C) The in vitro transcribed RNA is allowed to hybridize with the denatured aDNA library, and RNA:DNA hybrids are separated based on the binding of biotin to streptavidin. Data from L. Orlando, T. P. Gilbert and E. Willersley, 2015, Nature Reviews: Genetics, 16, pp. 395–405.

Even when the most careful laboratory techniques are used, some contamination of aDNA with modern material is inevitable. A variety of methods that exploit both genetic and evolutionary inference can be used to determine the degree of contamination. For example,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

when analyzing samples from Neandertals, the degree of contamination can be determined by the frequency of sequence reads that contain bases known to be derived in H. sapiens and ancestral in Neandertals, and those reads can be excluded as contaminants. In addition, the presence of Y chromosome markers in aDNA derived from a female and evidence for the presence of multiple X chromosome haplotypes in male remains are both indicators of contamination. To some extent, once massively parallel sequencing is completed, standard bioinformatic pipelines can be used to assemble the short reads into contigs. In the case of aDNA analysis, even with extensive enrichment for endogenous sequences, reads of sequences derived from other sources will be present and must be filtered out. In addition, statistical approaches can correct for misincorporations resulting from the damaged sites present in the original sample. Computer algorithms for addressing these problems are developing rapidly.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Ancient DNA is now widely used to study human evolution, and we will mention a few applications of it in this context later in the chapter. Molecular analysis of ancient material has other exciting possibilities as well: ■ RNA-seq (described in the Manipulating Genes and Genomes chapter) has been used to identify mRNA transcripts present in 700-year-old maize kernels (unfortunately, RNA molecules are not preserved in animal tissues). ■ DNA that is associated with nucleosomes is protected from postmortem damage, so DNA sequences associated with them are over-represented in aDNA samples. Thus, the depth of coverage of those sequences will be greater than of those in non-nucleosomal chromatin.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

■ If aDNA can be used to reconstruct entire genomes, might those genomes serve as the basis for “de-extinction” of extinct species? The completion of a draft sequence for the woolly mammoth has led to considerable speculation regarding this possibility. Obviously, going from a genome to an organism presents many technical and ethical challenges, and even if they can be overcome, genome assembly requires the availability of a reference sequence from a modern species (which rules out bringing back the dinosaurs, at least for now).

SUMMING UP ■ Analysis of aDNA is playing an increasingly important role in molecular evolutionary genetics. ■ Among the challenges associated with sequencing aDNA is the fact that it is usually degraded and can be contaminated by DNA from both ancient microbes and modern sources.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Sequencing of aDNA is typically performed by construction of a library that can be amplified by PCR and sequenced using high-throughput methods.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

19.3 Where Humans Fit on the Tree of Life We now turn to applying the tools of molecular evolution to understanding human evolution. Humans are primates, one of almost 20 orders of placental mammals. As long ago as 1863, Thomas Henry Huxley recognized that humans belong in a natural grouping with chimpanzees and gorillas based on morphological similarities. Hominoids are the group containing apes with small

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

bodies (gibbons, siamangs) and those with large bodies (humans, chimpanzees, bonobos, gorillas, orangutans). Since the 1960s, it has been recognized from immunological data that orangutans are genetically distinct from the other large-bodied hominoids. The relationship among humans, chimpanzees, and gorillas, however, was not determined for another 30 years.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Evidence That Humans Are Most Closely Related to Chimpanzees Chimpanzees are the species most closely related to humans. There are two types of chimpanzees—the common chimpanzee (Pan troglodytes) and the bonobo or pygmy chimpanzee (Pan paniscus). They are more closely related to each other than either is to any other primate species. Together the two species of Pan

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

form a clade (a group on an evolutionary tree that includes a common ancestor and all of its descendants), and both species are equally genetically close to humans. What startled some naturalists is that chimpanzees are not most closely related to gorillas (Gorilla gorilla), despite similarities between chimpanzees and gorillas in some distinctive features not found in humans, such as using knuckle-walking as a form of locomotion, similarities in cranial (pertaining to the head and skull) and postcranial (pertaining to the rest of the body exclusive of the skull) proportions, and pervasive body hair. Humans have so many unique features (upright gait, large brain relative to body size, language, culture, cooking, ability to run long distances, concealed ovulation, slow development, longevity, relative hairlessness) that many scientists assumed humans should be genetically very distinct from both chimpanzees and gorillas. Two main types of genetic evidence link humans and chimpanzees evolutionarily. The first is from DNA hybridization experiments, and the second is from DNA sequence studies of multiple, independently evolving genetic elements.

Similarities in Genomic DNA As discussed in the DNA Structure and Genetic Variation chapter, increasing temperature can cause the DNA strands in a duplex

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

molecule to come apart (denature). If the duplex consists of strands from different species, the temperature at which the strands separate depends on the extent of base pairing. Strands with a greater proportion of mismatches denature at a lower temperature than those that match better. The thermal stability of DNA duplexes formed by hybridizing the unique (nonrepetitive) DNA of pairs of species, therefore, provides a measure of how similar their genomic DNA is in nucleotide sequence.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

When such hybrid DNA molecules are formed between many different pairs of species of apes and humans, the hybrid requiring the most heat to denature is that between human and chimpanzee (FIGURE 19.8A). The greater thermal stability indicates that humans and chimpanzees have the greatest degree of DNA sequence similarity in their genomes, which is most easily explained if they share a common ancestor more recently than any other species pair in the experiment. Differences in thermal stability of hybrid DNA molecules can serve as a measure of distance between species, and a species tree can be deduced from such data, as illustrated for the primates in FIGURE 19.8B.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.8 Evidence for humans and chimpanzees as closest evolutionary relatives from DNA hybridization studies. (A) Melting temperatures Tm are shown for different double-stranded DNA molecules. Each of these is then expressed as a change in melting temperature ΔTm relative to the baseline melting temperature for the human double-stranded DNA molecule. (B) The ΔTm values are genetic distances used to infer the primate evolutionary tree. Data from A. Caccone and J. R. Powell. 1989, Evolution, 43, pp. 925–942.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

One limitation of DNA hybridization is that it gives an overall measure of sequence similarity without actually identifying any sequence differences. Which particular nucleotide substitutions shared by humans and chimpanzees are not present in gorillas? This issue is discussed next.

Analysis of Multiple Genetic Elements As an alternative to DNA hybridization, we can use gene trees to infer species trees, although we need to keep in mind two complicating factors regarding gene trees. First is the problem of incomplete lineage sorting, due to the presence of polymorphisms in related species that predate the division of the two species. Fortunately, incomplete lineage sorting happens only in a minority

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

of cases. Second, as is the case with any character, a particular sequence can be either ancestral or derived, depending on whether it is or is not identical to the corresponding sequence of an immediate ancestor. To classify bases in this way, we need to have a reasonable idea of which sequences were likely in the ancestor. This is not always possible to determine, especially when comparing rapidly evolving genes. Despite these limitations, the analysis of gene trees from many

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

genes in different regions of the genome that are evolving independently usually indicates the true species tree. Independence between the genes is important, because the analysis of tightly linked genes could yield gene trees different from the species tree if the linked genes were to share the same misleading pattern of incomplete sorting. If enough gene trees from independently evolving gene regions support a single tree, however, then the cumulative support increases confidence that the predominant tree is the true species tree (the actual phylogeny of the species). Studies of this type have found overwhelming statistical support for one tree—the one with humans and chimpanzees as most closely related, and gorillas as a more distantly related species (a sister group) to the human-chimpanzee clade. More definitive evidence came from another approach—an analysis of the distribution of highly repetitive Alu sequences. Recall that this family includes retroposon-derived repetitive sequences that are present in nearly 1 million copies in human (and other primate genomes). Importantly, most Alu sequences are transpositionally inactive. Thus, if we examine a particular location in two genomes and find an Alu sequence present in one but not the other, we can infer that the absence of the sequence is the ancestral state and the presence of it is the derived state. Mark Batzer and his colleagues examined the distribution of 101 Alu sequences in 8

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

primate species, and used parsimony analysis to infer a phylogeny. The results of their analysis are shown in FIGURE 19.9. Notice the levels of bootstrapping support: All but one of the nodes have 100 percent support (and the exceptional one could be explained by an unusual recombination event). Most importantly, these results firmly establish that humans and chimpanzees are, in fact, sister species.

FIGURE 19.9 Phylogeny of hominids based on analysis of the absence or presence of Alu sequences in the genomes studied where absence is the ancestral trait and presence is the derived trait. Numbers below the branches indicate the number of Alu insertions that have occurred on the branch; numbers above the branches indicate bootstrapping support. Reproduced from Salem et al, 2005, Proc. Nat'l. Acad. Sci, 100, pp. 12787-12791.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Copyright (2003) National Academy of Sciences, U.S.A.

Differences Between Human and Chimpanzee Genomes Chimpanzees differ from humans in many traits, including disease incidences, so the idea of comparing genomes to discover the genetic bases for such differences is compelling. To take one example, chimpanzees experimentally infected with the human immunodeficiency virus (HIV) do not progress to developing full-

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

blown cases of acquired immunodeficiency syndrome (AIDS), as most humans do in the absence of anti-retroviral therapy. Another example is that humans suffer from liver complications due to latestage hepatitis B or C, but chimpanzees do not. Questions about which genomic differences have accumulated over the last 6–8 million years of evolution between two such closely related hominoid species gave scientific impetus for sequencing the chimpanzee genome. The types and extent of differences between

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the human and chimpanzee genomes are summarized in FIGURE 19.10. Roughly 35 million base pairs of DNA differ between humans and chimpanzees because of simple nucleotide substitutions, which is equivalent to 1.2 percent of the genome. In addition, the species are distinguished by a surprising number of insertion or deletion events—that is, stretches of DNA present in one species but not in the other. The amount of DNA involved in these losses or gains is substantial: Approximately 45 million base pairs of DNA present in humans are not found in chimpanzees, and another 45 million base pairs of DNA present in chimpanzees are not in humans. Overall, counting both nucleotide substitutions and insertion or deletion events, humans and chimpanzees differ in about 4.2 percent of their genomes.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.10 Genomic differences between humans and chimpanzees. The genome of the common chimpanzee, Pan troglodytes, is used in this comparison. Data from the Chimpanzee Sequencing and Analysis Consortium. 2005, Nature, 437, pp. 69–87.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Most protein-coding genes show only subtle differences between the two species. The average gene has acquired only two amino acid changes since the species diverged. Approximately 30 percent of proteins are identical in humans and chimpanzees. Focusing on individual chromosomes, most autosomes show about the same amount of genetic differences between the two species. The sex chromosomes present a different story: The Y chromosome differs more between humans and chimpanzees than the autosomes, whereas the X chromosome differs less than the autosomes. Other genetic differences revealed by the chimpanzee genome pose puzzles, such as the discovery that a few alleles known to cause disease in humans are the wild-type allele in chimpanzees. These include alleles for common human diseases such as diabetes and coronary heart disease. For example, the PPARG

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

gene is important in lipid metabolism and development of fat cells; a PPARG allele with a proline at the 12th residue in the encoded amino acid is wild type in chimpanzees, but increases the risk of type 2 diabetes in humans. One possible explanation for this difference is that, in the common ancestor of humans and

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

chimpanzees, the disease-causing alleles were initially benign and widespread but subsequent physiological changes in humans (possibly in response to changes in ecology, environment, or behavior) modified the phenotypic effects of these alleles. Six large regions of the human genome are relatively invariant among all humans (showing far fewer than the usual numbers of polymorphisms), but they are as different between humans and chimpanzees as any other regions. It is likely that these chromosomal regions contain selectively advantageous genetic variants that arose as new mutations during hominin evolution. These variants gave rise to selective sweeps (see the Genes in Populations chapter), resulting in the low level of haplotype variation seen in modern human populations. The six nearly invariant genomic regions in humans are distributed on different chromosomes, and each is millions of base pairs in length (TABLE 19.1). While most of the regions contain several genes, one contains no known genes. Although these intriguing chromosomal regions may contain some of the genes responsible for the unique phenotype of all living humans, much work remains to be done in determining which genes or genetic elements are responsible. In one case, there is a hint: One such region in chromosome 4 has been associated with obesity in two different studies, suggesting that traits related to body weight or metabolism have been selected in humans since the divergence from our common ancestor with chimpanzees.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

TABLE 19.1 Six human genomic regions that underwent selective sweeps over the last 250,000 years Chromosome

Size

Genes

(Mb) 1

400

Fourteen known genes from ELAVL4 to GPX7

2

4.12

ARHGAP15 (partial), GTDC1, ZFHX1B

4

3.20

No known genes or predicted genes based on experimental evidence

8

2.63

UNC5D and FKSG2

12

4.32

Ten known genes from PAMCI to ATP2B1

22

4.07

Fifty-seven known genes from CARD10 to PMM1

Data from the Chimpanzee Sequencing and Analysis Consortium. 2005, Nature, 437, pp. 69–87.

SUMMING UP

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ DNA–DNA hybridization between human and chimpanzee DNA suggests a close evolutionary relationship between the two taxa. ■ Phylogenetic analysis based on sites of Alu sequence insertion has provided definitive evidence that humans and chimpanzees are sister taxa. ■ Sequence differences between humans and chimpanzees are subtle, and much remains to be learned about their functional significance.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

19.4 What Do the Genetic Differences Between Humans and Chimpanzees Mean? How do genotypic differences translate into phenotypic differences? Most traits are determined by multiple genetic and environmental factors that interact in complex ways, so the relation between genotype and phenotype for most traits is obscure. Some

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

of the observed genomic differences between humans and chimpanzees may contribute significantly to phenotypic differences, whereas others make little or no detectable difference. Humans differ in many significant ways from chimpanzees (FIGURE 19.11), and sifting among the genetic differences to find those responsible for phenotypic traits unique to humans is a subject of ongoing research.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.11 Some major phenotypic differences between human and chimpanzee (Pan troglodytes).

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(Left) © Potapov Alexander/Shutterstock; (Right) © Ralph Hutchings/Visuals Unlimited, Inc.

Molecular Adaptations Unique to Humans As we saw earlier in this chapter, the measure Ka/Ks—the ratio of nucleotide differences that change amino acids in a protein to those that do not—can be used to identify molecular adaptations in protein-coding genes. An unusually high ratio indicates that natural selection has actively promoted a change in an amino acid sequence between two species because the modified protein

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

confers some advantage in reproduction or survival to the organisms that carry it. Which molecular adaptations have been discovered when comparing human and chimpanzee genomes? The chimpanzee genome project analyzed sets of genes with related functions. The gene set with the highest Ka/Ks value that has diverged most rapidly between humans and chimpanzees is involved in epidermal differentiation (skin cell maturation). Genes relating to keratin

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

proteins in hair have also evolved quickly. Therefore, genes encoding features of our body's outer covering have been greatly modified over the last 6–8 million years. Other classes of rapidly diverging genes include those related to pregnancy, olfaction, immune function, and the regulation of physiological processes. Another class of genes showing a great difference between species consists of transcription factors. Human transcription factors have experienced 47 percent more amino acid changes than their chimpanzee counterparts. Many of these genes play crucial roles in early development. Although the types of genes that have undergone accelerated evolution along the human lineage have been identified, much research is still needed to understand how the particular genes of each type have a modified function in humans compared to chimpanzees and how they relate to biomedically relevant traits.

FOXP2: A Gene Related to Language Humans and chimpanzees differ in the FOXP2 gene, which has been associated with language abilities. A mutant form of the gene has been found in several individuals from three generations of a family with language problems and in a similarly affected but unrelated individual. The types of mutations found in FOXP2 are

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

diagrammed in FIGURE 19.12 The affected individuals showed two different types of problems—difficulty in articulating words with appropriate movements of the mouth and face, and difficulty with linguistic competence including some aspects of grammar, syntax, word comprehension, and language processing. In terms of its vocal quality, the speech of affected individuals has been described as “largely incomprehensible to the naive listener.” Nevertheless, these individuals can still talk, which implies that FOXP2 is not “the” gene for language but merely one of several genes related to the production of competent speech.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.12 Structure of part of the FOXP2 gene showing known human mutations that impair language as well as human differences from chimpanzees and other mammals. Data from S. E. Fisher, et al. 2003, Annu. Rev. Neurosci., 26, pp. 57–80.

Comparing DNA sequences of the FOXP2 gene across species revealed that it is among the top 5 percent of most highly conserved proteins. The particular amino acid that is mutated in the family with language problems does not vary in FOXP2 proteins in organisms spanning a large part of the tree of life from yeast to humans. Natural selection has allowed very few amino acid

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

changes in FOXP2 over millions of years, and the few changes that have occurred are concentrated along the lineage leading to humans. Over the last 6–8 million years, humans have acquired two amino acid changes in the FOXP2 protein (Figure 19.12). In contrast, only one amino acid change took place in the roughly 130 million years between the divergence of the mouse and the last common ancestor of humans and chimpanzees (FIGURE 19.13). Since the initial discovery that FOXP2 was mutated in some people with language difficulties, another FOXP2 mutation has been found

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

by screening other individuals with speech impairment. However, the two FOXP2 differences between humans and chimpanzees do not occur at amino acid positions known to affect language.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.13 Molecular evolution of the FOXP2 gene. Above the branches are numbers of inferred amino acid changes that have occurred. The four variable amino acid sites are shown for each species on the right.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Data from W. Enard, et al. 2002, Nature, 418, pp. 869-872.

FOXP2 is a transcription factor that is strongly expressed in the human fetal brain. The human FOXP2 mutations causing language impairment lead either to a reduced level of the encoded protein or to an amino acid change located in a crucial structural part of the protein. Lower than normal levels of FOXP2 protein or altered FOXP2 protein may cause the brain to develop abnormally in fetuses, resulting in speech and language problems. Mice genetically engineered to have two copies of the human FOXP2 gene show behavioral differences from wild-type mice. They show reduced exploratory behavior, and they differ in the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

ultrasonic vocalizations that play a social function mainly between mother and offspring. Physiologically, the mice produce less dopamine in the brain. Dopamine is a neurotransmitter affecting multiple aspects of behavior and cognition, including voluntary movement, mood, learning, attention, problem solving, and working memory. Anatomically, the FOXP2 transgenic mice show cellular changes in basal ganglia thought to increase the efficiency of the neural circuitry between basal ganglia and cortex. These circuits are known to control a variety of human traits including speech, visual memory, comprehension of words and sentences, the ability to do mental arithmetic, and cognitive flexibility. One view is that the changes in FOXP2 during hominin evolution not only facilitated the evolution of different aspects of language, but also laid the groundwork for human creativity.

Gene-Expression Differences Between

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Humans and Chimpanzees In the same way that biomedical researchers study gene expression differences between two human tissue types (for example, between cancerous and normal liver cells or between fetal and adult brain cells), evolutionary biologists study gene expression differences between tissues of closely related species. Microarray technology and RNA-seq (see the Manipulating Genes and Genomes chapter) make it possible to compare millions of gene transcripts from two different sources in a single experiment, revealing quantitative differences in the levels of mRNA from many genes simultaneously. The data from many individual genes can be summarized as a distance matrix of gene-expression differences, and the distance-matrix data can then be used to construct a tree. When the sources of mRNA are from different species, the evolutionary tree depicts overall gene expression divergence between the species.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Microarray technology has been used to compare gene expression of blood, liver, and brain tissues in humans, chimpanzees, and orangutans. The tree of brain-gene expression differences shows a relatively longer branch leading to humans than to chimpanzees from their common ancestor. In contrast, trees based on blood and liver gene expression show both branches to be approximately equal in length. This difference suggests that over the last 6–8 million years of evolution, genes expressed in the human brain have changed much more in level of expression than they have in the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

chimpanzee brain. As indicated in FIGURE 19.14, while genes expressed in the brain have undergone accelerated changes in expression in the human lineage, genes expressed in the liver show the greatest changes in expression overall. However, the rates of change in the expression of genes expressed in liver have been roughly equal along both lineages.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.14 Gene expression differences between humans and chimpanzees. Data from W. Enard et al. 2002, Science, 296, pp. 340-343 and W. -P. Hsieh, et al 2003,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Genetics, 165, pp. 747-757.

Transcription-factor genes have not only experienced a greater number of amino acid changes in humans than in chimpanzees, but their gene expression levels have also experienced a higher degree of change. Among this class of genes, most human changes relative to chimpanzees involve increases in gene expression. Because transcription factors help regulate other genes, this observation underscores the importance of gene regulatory changes in human evolution.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

SUMMING UP ■ Comparison of specific genes based on Ka/Ks ratios suggests that evolution of the outer body covering has been very rapid. ■ FOXP2, a transcription factor, evolves very slowly but has been implicated as a key factor in the evolution of speech in humans. ■ Transcriptome analysis shows that changes in levels of gene

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

expression have happened much more rapidly in the brains of humans than in those of chimpanzees.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

19.5 A Synopsis of Human Evolution Our knowledge of human phenotypic evolution comes mainly from studying the physical remains of early hominins—their fossilized bones (the focus of paleoanthropology) and their tools and other cultural remains (the focus of archaeology). Reconstructing human evolution is difficult because the remains are ancient (some as old as 7 million years) and scarce, and different types of evidence are

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

preserved differentially depending on their physical properties. Fossil teeth are more likely to be recovered than fossil wrist bones, and stone and bone tools last longer than woven baskets or the remains of ancient hearth fires. As a consequence, we have an incomplete and biased set of evidence upon which to reconstruct the past. Offsetting those problems, recent technological developments allow us to examine ancient remains in new ways. For instance, aspects of tooth development can now be studied using an imaging technique that does not destroy fossil material while it is being analyzed. Ancient DNA from bones can also be isolated and its DNA sequenced. This section offers a synopsis of human evolution, laying the groundwork for discussing molecular genetic evidence for different models of human evolution and particular traits unique to humans. To reconstruct the common ancestor of humans and chimpanzees, we need to be able to distinguish between ancestral and derived traits. The common ancestor of humans and chimpanzees was probably apelike because the likeliest phylogeny based on genetic evidence has gorillas as most closely related to the Homo/Pan clade and because gorillas resemble chimpanzees in many physical features. Applying the principle that evolution most likely follows a path with the fewest number of changes (parsimony) suggests an apelike ancestor, one resembling a mixture of gorilla and common chimpanzee features. This parsimonious reconstruction of the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

human–chimpanzee common ancestor seems reasonable because some of the unique features defining humans have multiple complicated and interrelated working parts, such as the ability to walk upright (requiring a remodeled pelvis, spine, and lower limbs) or our greatly enlarged brain (supporting language, complex reasoning, and culture). These complex features are unlikely to have evolved more than once or to have been lost in a lineage after they have evolved.

The Cast of Characters in Human Evolution

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Our synopsis is intended to capture some of the main features of hominin evolution. It is simplified and does not include all species or species still in dispute. The summary that follows is based on paleontological and archaeological evidence; genetic evidence is discussed later. In broad terms, hominins fall into two groups (FIGURE 19.15). The early hominins had small bodies and small brains, the size of ape brains. Unlike apes, they walked upright on two legs (bipedal locomotion), although they probably did not walk as efficiently as later hominins. These early forms were relatively primitive and more like apes in their brains and faces, but from the neck down they showed new, relatively derived features and were less apelike. These creatures lived exclusively in Africa in woodland habitats. They ate fruit when it was seasonally available and resorted to eating roots and tubers as second-choice foods when fruit was not available.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.15 A simplified tree of hominin evolution. Although not shown here, a third out-of-Africa dispersal must have occurred along the lineage leading to Neandertals, after Neandertals and modern humans split from their common ancestor, Homo heidelbergensis, in Africa. Hypothesized positions of Homo floresensis and Homo denisova are shown as dashed lines. Data from the Committee on the Earth System Context for Hominin Evolution. 2010. Understanding Climate's Influence on Human Evolution. National Academy of Sciences, Washington, DC.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Developmentally, these early hominins probably had relatively fast maturation rates—more like apes and unlike modern humans. This conclusion assumes that tooth development is representative of overall development. Early hominin first permanent molars erupted at 3–4 years as they do in chimpanzees, compared to about 6 years in modern humans. Throughout the roughly 5 million years of this first phase of early human evolution, teeth became bigger (except for male canine teeth, which became smaller) and tooth enamel became thicker. Adult males were 1.5–2 times larger than

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

adult females, and this degree of difference between the sexes (sexual dimorphism) is greater than that seen in living humans and tends toward a more apelike pattern. The earliest known species (Sahelanthropus tchadensis) lived 7 million years ago (Ma), while the latest known species of this early hominin group (Paranthropus boisei) lived until a little more than 1 Ma (Figure 19.15). The latter organisms are termed “megadont” because of their huge teeth. Sandwiched in time between these species were those of the genus Australopithecus, which are famous for having left footprints preserved in volcanic ash at Laetoli, Tanzania, at least 3.6 Ma. Australopithecus gave rise to the lineage that became the first species of Homo sometime between 2.5 Ma and 3 Ma ago in Africa. The term australopithecine refers collectively to members of Australopithecus and Paranthropus and reflects their relatedness. The second group of later hominins consists of species belonging to the genus Homo (Figure 19.15). Compared to the early species, the later hominins are distinguished by having bigger bodies, bigger brains, and longer legs. They were capable of running in addition to walking bipedally. This shift in overall body form presents a puzzle: Here we have a bigger animal with an enlarged and metabolically expensive brain requiring more energy

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

overall, yet able to maintain this phenotype with smaller teeth than its ancestors. This is a clear signal that early Homo was able to obtain food in a much more efficient manner than its ancestors did. The observed pattern in the hominin fossil record of increase in body and brain size coupled with a decrease in tooth size is real and demands an explanation, because it requires an innovation in energy use. The later hominins are sometimes divided into three subgroups:

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

transitional hominins, premodern Homo, and anatomically modern humans. Homo habilis was a transitional form (Figure 19.15). Its name—literally “handy man”—derives from the structure of its hand bones and its likely ability to manipulate objects more skillfully than its ancestors. The first stone tools are found at 2.6 Ma, but it is not clear which species made them. Among premodern Homo were Homo erectus and Homo neanderthalensis. H. erectus was a premodern species that first appeared around 1.9 Ma and existed until approximately 200,000 years ago. It resembled today's H. sapiens in body size and in body proportions, but H. erectus had a brain that was only twothirds as large. Its teeth and jaws were barely larger than those of a large-toothed living human. Adult females were not much smaller than adult males, which is in sharp contrast to the high degree of sexual dimorphism seen in australopithecines. Developmentally, H. erectus resembled apes and australopithecines in that the first permanent molars appeared at about 4–5 years of age, which is sooner than the 6 years required in modern humans, and it most likely had a diet like that of modern humans. Notably, H. erectus was the first hominin species to leave Africa, with this migration occurring approximately 1.8 Ma. Its fossils have been found in Africa, Europe, Southeast Asia, and South Asia. The relationship of H. erectus to H. sapiens, and especially the fate of the H. erectus

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

populations that left Africa and their descendants, are the subject of debate to which the molecular evidence can contribute. The descendants of H. erectus that are not members of our own species are often referred to as archaic humans. This is a very general term that covers a range of species within the genus Homo, including Homo heidelbergensis, generally recognized to be the common ancestor of Neandertals and modern humans (Figure 19.15). Fossil remains of H. heidelbergensis date from 600,000– 400,000 years ago and indicate that its brain was larger than that of H. erectus.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Homo neanderthalensis first appeared in the fossil record sometime between 200,000 and 400,000 years ago and went extinct roughly 28,000 years ago. In our synopsis, it is considered to be an archaic species, separate from H. sapiens, although it has been classified as a subspecies of our own, designated Homo sapiens neanderthalensis. (“Neandertal” is the modern spelling of “Neanderthal.”) Neandertals lived in Europe, the Near East, and Western Asia, but no examples are known from Africa. They arose from the descendants of H. erectus—more specifically, from those H. heidelbergensis that migrated out of Africa. Compared to H. erectus, Neandertals developed more slowly, and their first permanent molars took longer to erupt. Compared to anatomically modern humans, Neandertals were bigger boned and stockier, and they had a wider pelvis and bigger joints (FIGURE 19.16). Neandertals' chests were more barrel-shaped, and their faces were larger with substantial brow ridges, projecting front teeth, large noses, and receding chins. They were similar in brain size to anatomically modern humans living at the same time, but their bodies were somewhat heavier. Humans living today have smaller brains and smaller bodies compared to Neandertals and

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

ancestral modern humans. Neandertal limbs were relatively short, probably reflecting an adaptation to cold climates—a phenomenon observed in a number of other mammals. Neandertal skeletons reflect greater muscularity and endurance, compared to those of living humans, likely correlated with an energetically demanding

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

way of life. They hunted game effectively, made spears and sophisticated stone tools, and were the first hominins to bury their dead. The archaeological record suggests that Neandertals lived in only small social groups.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.16 Some major phenotypic differences between modern humans and Neandertals. (Left) © Ralph Hutchings/Visuals Unlimited; (Right) © Martin Shields/​Science Source.

Different kinds of evidence now support the hypothesis that two additional species of archaic hominins existed in Eurasia. On the Indonesian island of Flores, fossil remains of hominins have been found that, based on anatomic features, have been designated Homo floresiensis (FIGURE 19.17). There remains considerable uncertainty regarding their origin and time of extinction, but the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

most recent evidence suggests that the age of the most recent fossils is about 60,000 years, predating the arrival of H. sapiens in southeast Asia. Individuals of H. floresiensis were short in stature (adults were approximately 1 meter tall) and had cranial capacities comparable to that of H. erectus.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.17 An artist's impression of the appearance of Homo floresiensis compared with a modern human female. © Mauricio Anton/Science Source.

Even more remarkable has been the discovery of the Denisovans. In 2004, a 30,000- to 50,000-year-old finger bone was discovered in Denisova cave in southern Siberia. This very limited fossil evidence made anatomic analysis impossible, and as a result, its owner's status as a member of a different species remains an open question. However as we shall see, ancient DNA evidence

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

supports this species' designation as a sister group to H. neanderthalis. Our own species, Homo sapiens, also known as anatomically modern humans, first appeared in Africa a little less than 200,000 years ago. At this time, there were Neandertals and Denisovans in Europe and H. erectus (and possibly H. floresiensis) in Southeast Asia. Thus, as many as five hominin species existed on earth 200,000 years ago. Considerably later, around 60,000 years ago,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

some H. sapiens populations dispersed out of Africa to inhabit other parts of the Old World, in a second hominin diaspora from Africa; other H. sapiens populations remained in Africa (Figure 19.15). As H. sapiens spread across Eurasia, they began to live in the same geographical regions as Neandertals, Denisovans, and other archaic human populations. By 45,000 years ago, H. sapiens began to expand into Europe. Around this time or earlier, they reached as far as East Asia and Australia. In morphology, H. sapiens is distinct from other hominins in having a relatively flat face, including smaller brow ridges and a substantial forehead. Taking tooth development as indicative of overall development, the slower tooth development in H. sapiens suggests that modern humans mature at a somewhat slower rate than Neandertals did. The origin of H. sapiens may have coincided with the origin of symbolic language. By the time of the earliest known art (cave paintings, sculptures) at 30,000 to 60,000 years ago, symbolic expression and language were undoubtedly part of the human phenotype. Around 10,000 to 12,000 years ago, a significant shift in the culture and biology of modern humans occurred with the domestication of plants and animals, which allowed humans to live in even larger and more sedentary social groups than their predecessors. The

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

increase in group size eventually led to the development of complex civilizations and cities. Larger population sizes meant that infectious diseases could be sustained in populations; therefore, infectious diseases started to become a major selective force at the time when cultural and technological innovation was also sparked.

Models of Modern Human Origins A point of continuing debate is the relationship of Neandertals to

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

our own species: Were the Neandertals the nearest relative of H. sapiens that went extinct without hybridizing with us, or was there gene flow owing to hybridization between the two species? A similar question about hybridization concerns H. sapiens and H. erectus and their descendant archaic populations. FIGURE 19.18 outlines three main models (hypotheses) that have been proposed to explain how H. sapiens arose and how it is related to older hominins. All of the models incorporate the fact that H. erectus was the first hominin to leave Africa and inhabit Eurasia (the first “out of Africa” migration). The models differ in their interpretations of what happened to these geographically diverse H. erectus populations. In the multiregional model (Figure 19.18A), H. sapiens is seen as evolving in several geographical areas from different dispersed H. erectus populations in Africa and Eurasia at roughly the same time. This model assumes sufficient gene flow among the widely dispersed H. sapiens populations to maintain the biological integrity of the new species. It predicts that the last common ancestor of all living humans was likely to have been relatively ancient, perhaps as old as 1.8 million years, coinciding with the time of the last common ancestor of the H. erectus populations that left Africa.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.18 Three classical models of modern human origins. Red arrows represent the origin of Homo sapiens, which in the multiregional model (A) occurs multiple times in diverse locations but in the other two models occurs only once in Africa. Dotted lines indicate gene flow. In the hybridization model (C), gene flow represents admixture (hybridization) between archaic hominins and H. sapiens. T-bars in the African replacement model (B) represent extinctions of archaic hominins (descendants of H. erectus) and their replacement by H. sapiens without admixture.

In contrast, the African replacement model (Figure 19.18B) proposes that H. sapiens evolved only once in Africa and rather recently (approximately 200,000 years ago), and some of them then left Africa to inhabit other parts of the world (the second “out of Africa” event). The replacement model hypothesizes that when H. sapiens populations met archaic human populations descended from the first out-of-Africa migration, the two species did not interbreed; rather, modern humans replaced archaic humans and the latter went extinct. In this model, the last common ancestor of living human populations would have existed no more than 200,000 years ago.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

The third model, called the hybridization model (Figure 19.18C), proposes that H. sapiens evolved only once in Africa and, after its dispersal out of Africa, hybridization took place between modern humans and archaic humans as their populations made contact. The hybridization model predicts that the last common ancestor of modern humans would have been ancient because of the gene flow into our species from archaic humans.

SUMMING UP ■ Evolutionary analysis based on the fossil record requires distinguishing between ancestral and derived traits. ■ Hominins—members of the human lineage—are typically divided into early and later hominins. ■ Homo erectus and its descendants are often referred to as archaic humans. ■ Homo neanderthalis existed in Asia and Europe from 200,000 to 400,000 years ago, going extinct approximately 28,000 years ago.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Fossil and DNA evidence suggests the existence of two other species in the genus Homo that coexisted with Homo sapiens: H. floresiensis and H. denisova. ■ Three models for the origin of Homo sapiens have been proposed.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

19.6 Genetic Evidence for Modern Human Origins The sequence of nucleotides in DNA and amino acids in proteins can be used effectively to infer the evolutionary history of humans and their populations. Advances in sequencing technology have also made it possible to analyze ancient DNA samples—in particular, those extracted from remains of Neandertals, Denisovans, and prehistoric humans. The results support hybridization between the ancestors of some humans, Neandertals and Denisovans, since sequences from the latter two can be found in modern humans.

Tracing Human History Through

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Mitochondrial DNA Inferring ancestral history is easiest for DNA molecules that do not undergo recombination, such as mitochondrial DNA (mtDNA). These organelle genomes are also useful for tracing ancestry because in many species they evolve considerably more rapidly than nuclear genes. In humans, differences in mitochondrial DNA sequences accumulate among lineages according to a molecular clock at an average rate of approximately one change per mitochondrial lineage every 3800 years. For example, the people of Papua New Guinea have been relatively isolated genetically from the aboriginal people of Australia ever since these areas were colonized approximately 40,000 years ago and 30,000 years ago, respectively. The total time for evolution between the populations is, therefore, 70,000 years (40,000 years in Papua New Guinea and 30,000 years in Australia). If one nucleotide change accumulates every 3800 years on average, the expected number of differences in the mitochondrial DNA of modern-day Papua New Guineans and Australian aborigines is expected to be 18.4

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

nucleotides (calculated as 70,000/3800). Such calculations are very approximate because a molecular clock can be quite variable over any particular interval of time, and estimates of colonization times can change with new archaeological finds. (Since this mtDNA substitution rate was first estimated, new evidence has increased the colonization time of Australia to 50,000 years ago.) Nevertheless, this example shows how the rate of mtDNA evolution can be used to predict the number of differences between populations. In practice, the calculation is usually done the other way around, with the observed number of differences between pairs of populations being used to estimate the number of years since the populations have been geographically separated.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Probable historical relationships among human populations can be reconstructed based on differences in the nucleotide sequences of mtDNA. FIGURE 19.19 shows the inferred ancestral relationships of mtDNA based on the complete mtDNA sequences from 53 individuals representing human populations throughout the world and rooted with a chimpanzee mtDNA sequence. Because mtDNA undergoes no recombination, the tree is the genetic history of only a single DNA molecule; moreover, because mtDNA is maternally inherited, it is the genetic history of females. Nevertheless, the mtDNA tree shows several remarkable features: ■ African populations are first to diverge from the root of the human tree. ■ Much of the mtDNA diversity in African populations is not found among non-Africans; on average, the mtDNA of Africans shows about twice as much genetic variation as the mtDNA among non-Africans. ■ Three of the four major lineages of mtDNA are found only in sub-Saharan Africans (green).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

■ The age of the most recent common ancestor of all human mtDNA sequences (green circle) is approximately 170,000 ± 50,000 years. This common ancestor of all living human mitochondrial types has been called “mitochondrial Eve” in recognition of the fact that mitochondrial DNA is maternally inherited. Technically, the time of the genetic common ancestor is known as the coalescence time. ■ A restricted subset of mtDNA lineages is found among nonAfricans (violet and red); these sequences share a more recent common ancestor with six African mtDNAs (purple) than these African mtDNAs share with other Africans.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The age of the most recent common ancestor of the mtDNA lineage joining African and non-African populations (red dot) is approximately 50,000 ± 25,000 years. This is an upper estimate for the dispersal of modern humans out of Africa.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.19 Inferred evolutionary relationships among human mitochondrial DNA molecules based on analysis of the complete nucleotide sequence of mtDNA (16,000 base pairs) from 53 humans from geographically diverse populations. Data from M. Ingman, et al., 2000, Nature, 408, pp. 708–713.

These mtDNA results support the African replacement model, as the data are consistent with an African origin for modern humans based on tree topology and a common ancestor of modern humans

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

that was recent rather than one that was 1 or 2 million years old. There were also earlier hominin migrations out of Africa, but according to the mtDNA evidence, the descendants of these earlier migrants were apparently replaced by the people who came later. The age of the oldest fossils of Homo sapiens is 195,000 years ±

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

5000 years from Ethiopia. This time is consistent with 170,000 ± 50,000 years for the time of the most recent common ancestor of all of the mtDNA lineages in FIGURE 19.19. Even when a molecule evolves according to a molecular clock, paleontological or archaeological evidence is needed to anchor events in real time. Molecular data can give only relative branch lengths (e.g., one branch is twice as long as another). To convert relative times on a molecular tree into absolute dates expressed in years, at least one well-dated fossil associated with a node on the tree is needed to serve as a calibration point. In FIGURE 19.19, the calibration is based on a human–chimpanzee divergence time of 5 million years. Current evidence supports a value closer to 7 million years, so the dates in FIGURE 19.19 would be somewhat more accurate if multiplied by 7/5. A corrected coalescence time for mitochondrial Eve would then be 238,000 years ± 70,000 years. This somewhat older date is still consistent with the fossil evidence and with a recent African origin, as it is much less than the 1.8 million year estimate for the common ancestor of modern humans predicted by the multiregional model. Mitochondrial DNA studies of Neandertals also speak to the question of modern human origins. Those studies have found that Neandertal mtDNA is very different from that of modern humans. The several mtDNA types sequenced from Neandertals form a clade that is a sister group separate from the clade of modern human mtDNA types. Assuming a date for the human-chimpanzee divergence of 6 million years and applying the molecular clock, a

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

common genetic ancestor for modern humans and Neandertals is estimated to have existed at 466,000 years ago (within the range of 321,000–618,000 years with 95 percent confidence). This finding of separate clades for Neandertal mtDNAs and human mtDNAs suggests that no Neandertal mtDNA made its way into modern humans. The implication is that, if hybridization had occurred, it was not between Neandertal females and modern human males. A coalescence time represents the time of the last genetic common ancestor between two or more groups. Because it occurs prior to mutational divergence, the coalescence time necessarily comes before the time when the descendant groups of that ancestor separated from one another. This disparity explains why coalescence times estimated from genetic data are often older than times for population splitting based on archaeological or paleontological data. The population splitting time of Neandertals and modern humans was likely between 270,000 and 440,000 years ago.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The Neandertal Genome Using ancient DNA sequencing techniques, a draft sequence of the Neandertal genome has been assembled, using three fossils found in Vindija, Croatia. Subsequently, two additional genomes—one from the Caucasus and one from Siberia—have been characterized. The surprise finding is that living humans from Eurasia (sampled from France, China, and Papua New Guinea) have about 2–3 percent of their genome that descends from Neandertals, whereas the genomes of living humans from subSaharan Africa have no Neandertal DNA. As indicated in FIGURE 19.20, the direction of gene flow was from Neandertals into modern humans, not the other way around. Matings between the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

species were probably between Neandertal men and non-African human women, because modern human mtDNAs show no trace of a Neandertal contribution. Note that this result supports none of the three simple models for modern human origins presented in Figure 19.18. The data come closest to supporting the African replacement model, but with a small degree of hybridization between Neandertals and the common ancestral population of the H. sapiens who left Africa. Since all non-Africans show equal genomic contributions from Neandertals, no matter where they live,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the hybridization event must have happened at the very beginning of the H. sapiens migration out of Africa but before modern humans dispersed to various parts of the world, probably between 50,000 to 80,000 years ago. For the vast majority of our nuclear genome (96–99 percent), we last shared a common ancestor who lived recently in Africa.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.20 Hybridization between archaic hominins and modern humans occurred at least twice and involved different ancestors of modern human populations. These admixture events were revealed by analyses of genomic data from modern humans, Neandertals, and Denisovans. Melanesians are represented here by Papuans and Bougainville Islanders from New Guinea.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Data from D. Reich, et al. 2010, Nature, 468, pp. 1053–1060.

The Neandertal genome also gives us insight into our own genomes. Particular segments of the genomes of non-Africans can be identified that were likely of Neandertal origin. These genomic segments are distinctive in showing greater genetic diversity among non-Africans than among Africans. Gene regions that were likely favored by selection along the lineage leading to modern humans are revealed by comparing genomes of modern human, Neandertal, and chimpanzee. The most striking example of a selected region contains the THADA gene associated with type 2

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

diabetes, which suggests that modern human physiology has undergone selection for changes in energy metabolism. Several genes expressed in the skin also have undergone unique changes along the human lineage, highlighting skin as an organ that has undergone adaptive change. Genes associated with cognitive development have also been preferentially selected. One such gene is located in the region critical for Down syndrome, another has been associated with schizophrenia, and two others with autism. Collectively, these findings provide strong evidence that

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

modern human cognitive ability has been a selected trait on a recent evolutionary time scale. Studies of Neandertal nuclear DNA have also revealed the presence of the two unique amino acid changes in the languagerelated FOXP2 gene that are fixed in humans but not found in chimpanzees or other species; however, the presence of these mutations does not necessarily imply that Neandertals could speak. In terms of physical appearance, a distinctive and unique allelic variant of the melanocortin 1 receptor (MC1R) gene, known to regulate skin and hair color in many animals, was found in two Neandertals, and the nature of the variant allele indicates that some Neandertals may have had red hair and pale skin. No modern humans share this particular allele, but other mutations in the human gene produce a similar phenotype. Some Neandertals were also likely able to perceive the bitter taste of PTC (phenylthiocarbamide), as some modern humans do, based on variation at their TAS2R38 gene (see the Mendelian Genetics: The Principles of Segregation and Assortment chapter).

The Denisovan Genome The single preserved tooth found in Denisova cave was sufficient to construct a draft genomic sequence. The surprising outcome is that

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Denisovans, although closer to Neandertals than to modern humans, are genetically distinct from Neandertals. As indicated in Figure 19.20, they are a sister group to Neandertals. Denisovans may represent a new hominin species previously unknown to paleontologists, achieving that status through characterization of their ancient DNA. When Homo sapiens left Africa, therefore, three archaic groups already inhabited Eurasia—Neandertals, Homo erectus, and the Denisovans.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Among modern humans, a small amount of Denisovan DNA has been found in East Asians, South Asians, and Native Americans. Surprisingly, that percentage is higher (4–6 percent) in populations from Oceania (Papua New Guinea and Melanesia; FIGURE 19.20). The origin of Denisovan DNA in non-Melanesians remains unclear. Gene flow between Denisovans and the ancestors of Oceanians likely occurred after the Neandertal DNA was hybridized with that of the human ancestor of non-Africans, but before the ancestors of Melanesians left continental Asia (no later than 50,000 years ago) to colonize the islands northeast of Australia and Papua New Guinea. At first glance, Denisovan mitochondrial DNA poses a puzzle because it is very different from that of all known Neandertals and modern humans. On the mtDNA tree, Denisovans are an outgroup to humans and Neandertals, unlike on the tree based on their nuclear genome. Denisovans appear to be more archaic in their mtDNA than in their nuclear genome. The mtDNA coalescence time for Denisovans, Neandertals, and humans is roughly 1 million years, while the mitochondrial coalescence time for Neandertals and modern humans is half as long, at 500,000 years. It is possible that Denisovans may have retained their archaic mtDNA type while Neandertals and living Melanesians lost any traces of it through incomplete lineage sorting. Another possibility is that Denisovans

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

themselves hybridized with some other as yet unknown archaic hominin. The discovery of new hominin fossil specimens (and their genomes) will help clarify these alternatives, while adding complexity to our knowledge of hominin evolution that we have long suspected was incomplete.

SUMMING UP ■ Ancient DNA analysis has found evidence of hybridization among Homo sapiens, Neandertals, and Denisovans. ■ Mitochondrial DNA analysis supports the African replacement model of human origins. ■ Numerous instances of Neandertal genes that may have subject to positive selection in humans have been reported. ■ The Denisovan genome is distinct from that of Neandertals.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Genes with Denisovan origins have been found in several human populations, most notably in Oceania.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

19.7 Measuring Human Diversity We introduced the 1000 Genomes Project in the Genes, Genomes, and Genetic Analysis chapter. This project has sequenced 2504 diploid genomes from 26 worldwide populations, which have been divided into 5 “superpopulations.” Based on these sequences, more than 80 million SNPs have been identified. In addition, using both pedigree and statistical analysis, the genotypes have been “phased,” meaning that alleles of SNPs in them have been assigned to either maternal or paternal chromosomes. Adjacent SNPs in a region of a chromosome tend to be inherited together as a block and, therefore, to be in linkage disequilibrium. Specifically, only about 500,000 SNPs need be studied to detect disease associations, because adjacent SNPs tend to be inherited as “haplotype blocks.” In this section, we will see how this kind of genomic data can be used to learn about genetic history of and relationships among different human populations.

Tracing Human History with Genetic

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Markers SNP data can be used to understand genetic relatedness among human populations. In one large study, 650,000 SNPs were examined in 938 unrelated individuals from 51 human populations worldwide. Approximately 97 percent of the SNPs were autosomal, 2 percent X-linked, and 1 percent in the Y chromosome or mitochondrial DNA. The subjects lived in places where their grandparents and ancestors had lived, with the researchers purposefully excluding people who had immigrated, and the subjects were analyzed as individuals and not initially considered as members of particular groups. FIGURE 19.21 shows the tree summarizing the genetic relatedness among these individuals, with several features being especially noteworthy:

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

■ The SNP-based tree shows greater resolution among human populations than the mtDNA tree in Figure 19.19. The greater resolution is not surprising given that over 40 times more data sampled across every chromosome were used to construct the SNP tree. ■ Individuals from the same continent or continental subregion usually appear together in a clade or in closely related clades. Non-Africans, in particular, tend to resemble their neighbors genetically. Some African groups, however, are more closely related to non-African groups than to other Africans, as seen in the mtDNA tree. ■ Clades of sub-Saharan Africans appear near the root of the tree, followed (in order) by populations from the Middle East, Europe, Central and South Asia, Oceania, East Asia, and the Americas. Populations that are geographically more distant from Africa tend to be farther from the root of the tree, and the order of their appearance on the tree is roughly consistent with the pattern of human expansion out of Africa reconstructed from paleontological and archaeological evidence.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Human populations have fewer heterozygous genes, indicating reduced diversity, according to their distance from a reference point in Africa.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.21 Inferred evolutionary tree of modern human populations based on more than 650,000 SNPs from 938 individuals from 51 populations. Data from J. Z. Li, et al. 2008, Science, 319, pp. 1100–1104.

The tree in FIGURE 19.21, based primarily on nuclear DNA data obtained from modern humans, supports an African origin for modern humans. The pattern of reduced heterozygosity has been interpreted as a model of serial colonization, in which successive

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

colonies are created from a subset of people from the previous colony (FIGURE 19.22). In this model, the first population of modern humans that left Africa was a small subsample of the original population with less genetic diversity. Over time, the small colony grew, and eventually another small subsample of people budded off and colonized a second geographical region even more distant from Africa. This second founding group had even less genetic diversity than the first. This process continued, creating a series of populations that were progressively more distant from

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Africa and less heterozygous with each step in the chain. FIGURE 19.23 presents one scenario describing how modern Homo sapiens colonized the world, based on a combination of genetic, paleontological, and archaeological evidence.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.22 The serial colonization model for modern human dispersal out of Africa is characterized by a series of founding events in which only a small subsample of each source population gives rise to a new population (colony) in a geographically separate location. Each founding event represents a genetic bottleneck in which genetic diversity of the new colony is reduced compared to diversity of the source population.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.23 One possible scenario for the migrations of modern Homo sapiens based on genetic, paleontological, linguistic, and archaeological data. The date 200,000 years (200 ka) represents an estimate of the first appearance of Homo sapiens in southern Africa. Humans began to disperse out of Africa around 60 ka. Not shown are the two known hybridization events with Neandertals and Denisovans. Hybridization with Neandertals likely took place immediately after modern humans first left Africa, when they reached the Near East and before they dispersed to other parts of the globe. The time when humans reached different continents and which routes they took are still under debate, and these may change as new fossil evidence is discovered.

Recent findings from ancient DNA suggest that, while serial colonization events have indeed occurred, subsequent interbreeding, or admixture, among individuals played a critical role, even in prehistoric times. For example, consider the relationship of Native Americans with East Asians. Under the serial colonization model, we would expect the modern Native American gene pool (excluding that portion resulting from recent contact with European Americans and African Americans) to comprise a subset

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

of that of modern Siberians, and the modern Siberian gene pool to be similar to that of the ancestral population from which Native Americans arose. In fact, when the genomes of two 24,000-yearold Siberian fossils were characterized, they were found to be quite different from modern Siberians and more closely related to Native

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Americans (and, surprisingly, to modern individuals in Western Asia). Modern Siberians appear to have originated as a result of migration from East Asia that occurred subsequent to the peopling of North America. FIGURE 19.24 depicts a technique that is commonly used to analyze and visualize genomic variation, in this case with the same populations shown in Figure 19.21. In this computer-based analysis, a certain number of ancestral groups is specified (seven, in this case), and then marker genotypes within individuals are assigned to each of the groups in such a way as to minimize deviations from Hardy Weinberg and linkage equilibria (see the Genes in Populations chapter). Thus, Figure 19.24 consists of a series of 938 vertical lines, each one representing the multilocus genotype of an individual in the study. The colors represent the portions of the genome derived from each of the seven ancestral populations. As we can see, particularly in the Middle East, many modern genomes are highly admixed, suggesting shared ancestry with individuals from a wide range of geographic areas. Even in areas like Northern Europe, where little evidence for admixture is evident from the 1000 Genomes Project data, when more individuals are genotyped and ancient DNA analysis is included, the picture becomes much more complicated.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.24 Analysis of geographic differentiation and admixture in human populations. Based on the assumption of there being seven ancestral populations (see Figure 19.23), computer analysis was done to assign the genotypes each of 938 individuals into those populations. Each genotype is a vertical line; the colors of the lines indicate the origins of the corresponding fractions of those genomes. Reproduced from J. Z. Li, et al. 2008, Science, 319, pp. 1100–1104. Reprinted with

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

permission from AAAS.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

The Apportionment of Within-Group and Between-Group Variation The human species is hierarchically structured. At the lowest level are individual populations, above them are clusters of populations within any given geographical region, and finally the hierarchy is topped with the groups of clustered populations from major continental regions that collectively make up our species. When

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

human genetic variation based on more than 1 million autosomal SNPs in the 1000 Genomes Project data is partitioned among these levels, 87 percent of the genetic variation is contained within populations, 1 percent among populations within superpopulations, and 12 percent is contained among superpopulations. Thus, the overwhelming majority of genetic variation within our species is the variation observed at the very lowest hierarchical level—that is, within populations. Variation in the X chromosome (based on more than 32,000 SNPs) yields a similar picture: 80 percent of the variation is found in the within-population component, 2 percent among populations within geographical regions, and 18 percent among geographical regions. These genome-wide studies showing that most human genetic variation occurs within local populations demonstrate why the population is by far the most important focus for the study of human genetic diversity.

Tracing Human History Through the Y Chromosome Because the Y chromosome does not undergo recombination along most of its length, genetic markers in the Y chromosome are completely linked and so remain together as the chromosome is transmitted from generation to generation. The genetic relatedness among Y chromosomes can, therefore, be traced: Chromosomes

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

that are closely related will share more alleles along their length than will more distantly related chromosomes and, hence, will have closely related haplotypes. For many genealogical studies of the Y chromosome, simple-sequence repeat (SSR) polymorphisms are convenient targets of study because of their relatively high rate of mutation due to replication errors and the large number of alleles. The logic is that Y chromosomes with haplotypes that share alleles at each of 20–30 SSRs across the chromosome must have descended from the same ancestral Y chromosome in the very

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

recent past. For haplotypes differing at a single locus, the genetic relationship is less close; for those differing at two loci, it is even less close; and so forth. This simple logic is the basis of tracing population history through Y-chromosome polymorphisms. Haplotypes that share many alleles have a more recent common ancestral Y chromosome than haplotypes that share fewer alleles. Furthermore, because the rate of SSR mutation can be estimated, the time at which the ancestral chromosome existed can be deduced. This reasoning forms the basis of the estimate that the most recent common ancestor of all extant human Y chromosomes existed 50,000–150,000 years ago. Such estimates are not highly precise, and require many assumptions to be made. Nevertheless, much can be learned about human population history through studies of the Y chromosome, as illustrated in the following example.

THE CUTTING EDGE: The Peopling of Western Europe

L

ook again at FIGURE 19.21 and focus on the “European” clade. Recall that these relationships were based on

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

comparisons of SNP genomes in modern Europeans, and they seem to imply that these humans originated as a result of migration from the Middle East. But what information do we have about the history of humans in Europe that might enrich our understanding of the species history in the region? And what if we could do population genomics on ancient populations in Europe and elsewhere?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Traditionally, evidence regarding prehistoric human populations comes from physical remains (fossils) and associated cultural artifacts. We know, for example, that the oldest European remains of modern Homo sapiens date back to approximately 45,000 years ago, and that humans have persisted in Europe continuously since then. We also know, based on cultural artifacts, that the “Neolithic Transition”—from a hunter–gatherer lifestyle to an agrarian one—occurred in the Middle East approximately 10,000 to 12,000 years ago and spread into Europe over the next several thousand years, following two routes—one along the Mediterranean coast and one up the Danube River (FIGURE A). But what can genetics tell us?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE A Proposed routes of gene flow into Western Europe in the Neolithic and Post-Neolithic periods. Shown in red are those from the Middle East about 7500 BCE, leading to admixture between migrant agrarian peoples and the resident hunter– gathers. Shown in blue is gene flow from the steppes of Asia about 4000 BCE, which is also suggested as an origin of IndoEuropean Languages in Europe. Proposed routes of gene flow into Western Europe in the Neolithic and Post-Neolithic periods. Shown in red are those from the Middle East about 7500 BCE, leading to admixture between migrant agrarian peoples and the resident hunter–gathers. Shown in blue is gene flow from the steppes of Asia about 4000 BCE, which is also suggested as an origin of Indo-European Languages in Europe.

Sequencing ancient DNA, while it yields informative results, is technically challenging. However, as shown in FIGURE 19.21, much can be learned from robust SNP genotyping, as opposed to complete genome sequencing. In 2015, a large group of researchers from several labs reported doing exactly this— genotyping 69 ancient samples from across Europe and Central Asia, which dated to 1000 to 5000 BC (determined by radiocarbon dating). What made this large-scale analysis possible was the use of oligonucleotides specific for known SNPs as hybridization “baits,” which were used to enrich aDNA libraries for regions containing SNPs. The approach was

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

successful: In total, the 69 samples were successfully genotyped for an average of 212,000 SNPs. The major results are summarized in FIGURE B and placed in a historical and geographical context in Figure A. In Figure B, each bar represents samples from a population of individuals, with their inferred ancestry being shown. Using these data, the authors were able to draw a series of important conclusions: ■ The earliest samples, from Hungary in south central Europe, were almost entirely Neolithic in character, consistent with other data suggesting a migration of agrarians from the Middle East around 7500 BCE. ■ As time progressed (until about 4500 BCE), there was admixture between individuals with Neolithic ancestry and the hunter–gatherers who had occupied Europe in the preNeolithic years, such that the fraction of individual genomes derived from them increased.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ At approximately 4500 BCE, there was a rapid influx of genes from the steppe region of modern-day Russia, such that in some populations there is evidence of little or no ancestry derived from either the early Neolithic migrants or the more ancient huntergatherers.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE B Estimation of admixture of ancient populations surveyed. Note the dramatic increase in Yamnaya (steppe) ancestry found in German populations from approximately 4000 years ago.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Data from Haak et al., 2015, Nature, 522, pp. 207–211.

Thus, in this case, the serial colonization model serves as a starting point, but a more complete picture requires that we incorporate admixture of colonists with existing populations, as well as additional waves of expansion and admixture that occurred at later times. These data, in turn, raise intriguing questions about the origins of Indo-European languages in Europe. Linguists have hypothesized that they have their roots in either Anatolia in present-day Turkey, spreading to Europe during the Neolithic, or later, from the steppes north of the Black and Caspian Seas, with the advent of wheeled vehicles. The abrupt genetic transition that occurred around 4500 BCE provides support for the second hypothesis.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

References Haak, W., Laziridis, I., Patterson, N., Rohland, N., Mallik, S., Llamas, B…. Reich, D. (2015). Massive migration from the steppe was a source for Indo-European languages in Europe. Nature, 522, 207–211.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

In the thirteenth century, the Mongol Empire developed as the largest land empire that history has known. At its maximum extent, it stretched from China to Russia through to the Middle East and then into Eastern Europe. The founder was born around 1162 and given the name Temujin. As a young man, he organized a confederation of tribes, who around 1200 took to their small Mongolian ponies equipped with high wooden saddles and stirrups, and armed with bow and arrow began to conquer their neighbors. Soon thereafter, Temujin adopted the name Genghis Khan, which means “Universal Ruler.” He was often merciless, exterminating the men and boys of rebellious cities and kidnapping the women and girls. In answer to a question about the source of happiness, he is reputed to have said, “The greatest happiness is to vanquish your enemies, to chase them before you, to rob them of their wealth, to see those dear to them bathed in tears, to clasp to your bosom their wives and daughters.” Through their multiple wives, concubines, and innumerable unrecorded sexual conquests, Genghis Khan and his descendants were very prolific. His eldest son, Tushi, had 40 acknowledged sons, and his grandson Kubilai Khan (under whom the Mongol Empire reached its maximum extent) had 22 acknowledged sons. Although the legacy of Genghis Khan is well recorded in history, it was hardly expected to show up in studies of the Y chromosome. Surprisingly, though, genotyping studies of 32 markers along the Y chromosome of 2123 men sampled from a large region throughout

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Asia yielded the remarkable result in FIGURE 19.25. Each circle represents a population sample, with its area proportional to the sample size. The red sectors denote the relative frequencies of each group of nearly identical Y chromosome haplotypes, whereas the white sectors represent the relative frequencies of other haplotypes that are genetically much more diverse. The most recent common ancestor of the closely related haplotypes is estimated as existing 1000 ± 300 years ago. Furthermore, the geographical region in which the closely related haplotypes cluster

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

is included largely within the Mongol Empire (shading). The sole exception is population 10, composed of the ethnic Hazara of Pakistan. This population provides a clue to the origin of the closely related Y chromosomes, because the Hazara consider themselves to be of Mongol origin, and many claim to be direct male-line descendants of Genghis Khan. Whatever their origin, the closely related Y chromosomes are found in approximately 8 percent of the males throughout a large region of Asia (populations 1–16).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FIGURE 19.25 Distribution of Y-chromosome haplotypes (red), presumed to have descended from Genghis Khan or his close male relatives, among populations near and bordering the ancient Mongol Empire. The specific population groups are (1) Mongolian, (2) Han [Gansu], (3) Chinese Kazak, (4) Han [Xinjiang], (5) Xibe, (6) Uyghur, (7) Kyrgyz, (8) Kazak, (9) Uzbek, (10) Hazara, (11) Hezhe, (12) Daur, (13) Ewenki, (14) Han [Inner Mongolian], (15) Inner Mongolian, (16) Manchu, (17) Oroqen, (18) Han [Heilongjiang], (19) Chinese Korean, (20) Korean, (21) Japanese, (22) Shezu, (23) Han [Guangdong], (24) Yaozu [Liannan], (25) Lizu, (26) Buyi, (27) Yaozu [Bama], (28) Huizu, (29) Han [Sichuan], (30) Hani (31) Qiangzu, (32) Chinese Uyghur, (33) Tibetan, (34) Burusho, (35) Balti, (36) Kalash, (37) Tajik, (38) Baloch, (39) Parsi, (40) Makrani Negroid, (41) Makrani Baloch, (42) Brahui, (43) Turkmen, (44) Kurd, (45) Azeni, (46) Armenian, (47) Lezgi, (48) Georgian, (49) Ossetian, and (50) Svan. Data from T. Zerjal, 2003, Am. J. Hum. Genet., 72, pp. 717–721.

Direct proof of the connection with Genghis Khan could, in principle, be obtained by determining the haplotype of the Y chromosome in

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

material recovered from his grave. He died in 1227 from injuries sustained in a fall from a horse, and he is said to have been buried in secret at a concealed site near his birthplace.

SUMMING UP ■ Analysis of approximately 650,000 SNPs in various populations has led to an inferred phylogeny with much greater resolution than that obtained from mitochondrial DNA. ■ This analysis supports a model of serial colonization, beginning in Africa and extending to other parts of the world. ■ While genetic differentiation of modern human populations can be observed, it involves only about 12 percent of the genome.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ The Y chromosome can be used to trace paternal lineages, such as that descending from Genghis Khan.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

19.8 Genetic Adaptations Unique to Humans As human populations colonized different regions of the world (FIGURE 19.23), they faced novel environments with different climates and ecosystems, and humans adapted in response to these physiological challenges. Exposure to local parasites and other environmentally restricted diseases imposed selective forces on the human immune system differing by geography. New environments also provided opportunities for exploiting new local food resources while forfeiting some staple foods. We have already seen how domestication of dairy animals led to selection for the lactase persistence phenotype in Europe and Africa (the “Molecular Signatures of Selection” section in the Genes in Populations chapter). Culture played an important role, too—for example, in developing new types of tools and foraging strategies. In the following sections, we examine a few of the genetic adaptations that have so far been discovered.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Amylase and Dietary Starch With the development of agriculture, human diets changed as populations took advantage of local plants and turned them into cultivated food sources. Plant species such as wheat in the Middle East, rice in Asia, and corn in the Americas became domesticated. Plant foods differ in the amount of dietary starch. Domesticated plants generally contain more starch than their wild ancestors, and human groups differ in the degree to which they rely on starchy plants. High-starch diets are characteristic of almost all agriculturalists and hunter–gatherers who live under arid conditions and rely on roots and tubers. Low-starch diets are the usual fare among pastoralists, among hunter–gatherers in wet conditions such

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

as tropical forests, and among arctic dwellers with diets rich in meat or fish. Different diets are likely to have exerted different selection pressures on human physiology since the advent of agriculture.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Digestive enzymes are, therefore, prime candidates as potential targets of selection among human populations. Amylase is an enzyme present in saliva and the pancreas that hydrolyzes starch. Salivary amylase performs the first step in starch digestion as food is chewed. Studies of copy-number variation across the human genome (discussed the DNA Structure and Genetic Variation chapter) revealed that humans differ in the number of genes for amylase present in their genomes. As indicated in FIGURE 19.26, examination of populations with low-starch diets or high-starch diets showed that the copy-number variation correlates with dietary starch (an average of 5.4 copies in low-starch populations versus an average of 6.7 copies in high-starch populations). Since populations with high-starch and low-starch diets can live near each other, geographical ancestry is not an alternative explanation for the pattern. Furthermore, chimpanzees have only one copy of amylase per haploid genome and bonobos have what appear to be two nonfunctional copies; hence, the likeliest ancestral state for hominins is to have had a low gene-copy number. This suggests that the number of amylase gene copies has probably increased during human evolution. Although the advent of agriculture may have promoted an increase in amylase gene copies, it is also possible that selection could have begun when hominins shifted in habitat from woodlands and forests to more arid savannahs and came to rely on starchy roots and tubers.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

FIGURE 19.26 Amylase copy number varies with the amount of starch in human diets. Data from G. H. Perry et al. 2007, Nature Genetics, 39, pp. 1256–1260.

ROOTS OF DISCOVERY Starch Contrast Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

George H. Perry1,2, Nathaniel J. Dominy3, Katrina G. Claw1, Arthur S. Lee 2, Heike Fiegler4, Richard Redon4, John Werner, and six other authors (2007) 1. Arizona State University, Tempe, Arizona. 2. Brigham and Women's Hospital, Boston, Massachusetts. 3. University of California, Santa Cruz, California. 4. The Wellcome Trust Sanger Institute, Hinxton, United Kingdom.

Diet and the Evolution of Human Amylase Gene Copy Number Variation

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Evolution of increased enzyme activity can occur through regulatory mutations that increase transcription of a single gene or through increases in gene copy number. This study reports a strong correlation between the amount of starch in the diets of human populations and the number of copies of a gene encoding salivary amylase, a starch-degrading enzyme. Humans are not alone among primates in having evolved higher amylase activity. A group of Old World monkeys called cercopithecines, which includes macaques and mangabeys, produces even more salivary amylase than humans. Cercopithecines are unique among primates in storing starchy foods, such as the seeds of unripe fruits, in a cheek pouch, and it is a plausible hypothesis that the increased amylase facilitates digestion of the starch. It is not known whether the increased amylase production in cercopithecines is due to copy-number variation or to some other mechanism.

“We favor a model in which AMY1 copy number has been subject to positive or directional selection in at least some high-

”

starch populations …

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

First, we can consider the variation in starch content in diets of different human populations. A distinction can be made between “high-starch” populations for which starchy food resources comprise a substantial portion of the diet and “low-starch” populations with traditional diets that incorporate relatively few starchy foods [but] instead emphasize proteinaceous resources (for example, meats and blood) and simple saccharides (for example, from fruit, honey, and milk)…. Then the genomes of individuals from those populations can be analyzed to determine the copy number of the AMY1 gene, which encodes salivary amylase.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

We estimated AMY1 copy number in three high-starch and four low-starch population samples…. Notably, the proportion of individuals from the combined high-starch sample with at least six AMY1 copies (70 percent) was nearly two times greater than that for low-starch populations (37 percent)…. One again, however, the scientific maxim that “correlation does not necessarily imply causation” applies here. Are these differences truly the result of evolutionary adaptation to starch in the diet, or could something like genetic drift be responsible? The authors note that AMY1 copy number correlates more highly with dietary starch than with the geographic distribution of the populations study. They conclude We favor a model in which AMY1 copy number has been subject to positive or directional selection in at least some high-starch populations but has evolved neutrally (that is, through genetic drift) in low-starch populations…. Comparisons with other great apes suggest that AMY1 copy number was probably gained in the human lineage… . The initial human-specic increase in AMY1 copy number may have been coincident with a dietary shift early in hominin evolutionary history.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Source: Green, R. E. et. al. (2010). Science, 328 (5979): 710-722.

Adaptation to Parasites and Disease Since the advent of agriculture approximately 10,000 years ago and the resultant increase in population density, infectious diseases have been a health burden on human populations. Natural selection has, therefore, favored alleles that protect against infectious disease, and no disease illustrates this point more clearly than malaria. Multiple mutations in different genes have been identified that confer partial protection against malaria, and all of them have

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

arisen in the human population less than 10,000 years ago. The list of these changes includes the mutations in TABLE 19.2. ■ Mutations changing the amino acid sequence of betahemoglobin. The sickle-cell allele HbS resulting in the amino

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

acid change Glu6Val was discussed in the Genes in Populations chapter. This allele is relatively common across a broad swath of central Africa as well as in other geographical regions where malaria was or still is endemic. The diversity of HbS alleles suggests that this mutation arose independently at least four times in Africa. Another allele, HbC, has a mutation in the same codon producing Glu6Lys. The highest allele frequency of HbC is about 0.2, and it occurs in a restricted region of central West Africa. The HbC allele is less than 5000 years old, and both heterozygous and homozygous genotypes enjoy a protective effect against malaria and are associated with only minor anemia. Still another allele, HbE, has the amino acid replacement Glu26Lys. Heterozygous genotypes have high resistance to malaria and no adverse health effects, whereas homozygous genotypes have less protection as well as mild anemia. The HbE allele reaches high frequencies in eastern India, southeast Asia, and Myanmar, and is especially common in northern Thailand and Cambodia, where 70 percent of people are heterozygote carriers. The HbE allele is estimated to have arisen only about 1000–4000 years ago. ■ Mutations changing the amount of beta-hemoglobin. The term “thalassemia” refers to a group of diseases in which the globin chains are produced in reduced amounts. Betathalassemia can result from mutations in promoter or enhancer regions that reduce the level of transcription, but more commonly it results from deletion of all or part of the betaglobin gene. Heterozygous genotypes have some protection from malaria with few adverse symptoms, whereas

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

homozygous genotypes have severe, life-threatening anemia requiring regular blood transfusions or bone-marrow transplant. ■ Mutations changing the amount of alpha-hemoglobin. Chromosome 16 normally carries two copies of the alphaglobin gene. Deletion of one or both of these copies is associated with alpha-thalassemia. In heterozygous genotypes, these deletions reduce the amount of hemoglobin and confer some protection against malaria; in contrast, in

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

homozygous genotypes, they can cause serious disease. Individuals with three copies of the alpha-globin gene are clinically normal, but those with two or fewer copies have anemia whose severity depends on copy number. Fetuses missing all four copies die in utero. ■ Deficiency of glucose-6-phosphate dehydrogenase (G6PD). Stability of the red blood cell requires the enzyme G6PD. When this enzyme's activity is reduced, cellular stress can cause the red blood cells to burst. Infection of red blood cells with the malaria parasite is a form of cellular stress, and certain alleles of the X-linked gene for G6PD that reduce activity confer protection against severe infection. The most common mutant allele in Africa has two amino acid replacements (Val68Met and Asn142Asp) that reduce G6PD activity by about 88 percent. The allele is approximately 4000– 12,000 years old, and in heterozygous females and carrier males, the allele decreases the risk of severe infection by about 50 percent. Another, more recent allele dates to approximately 2000–7000 years ago and has a Ser188Phe replacement that decreases enzyme activity by about 97 percent. Most individuals with these alleles have no clinical symptoms; however, certain environmental factors can lead to red blood cell destruction and anemia. The most common environmental triggers are aspirin, nonsteroidal anti-

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

inflammatory drugs, virus or bacterial infection, ingestion of fava beans, and exposure to the chemical in moth crystals. TABLE 19.2 Genetic adaptations that protect against malaria Gene

Major types of mutations

Beta-globin

HbS (Glu6Val), HbC (Glu6Lys), HbE (Glu26Lys)

Beta-globin

Partial or complete deletion (beta-thalassemia)

Alpha-globin

Deletion of one or both duplicates (alpha-thalassemia)

G6PD

A– (Val68Met + Asn142Asp), Med (Ser188Phe)

Malaria is hardly unique in having elicited genetic adaptations for disease resistance. Another example is found in two alleles for a blood apolipoprotein found only in regions of sub-Saharan Africa in which sleeping sickness is endemic. This disease is caused by a protozoan parasite transmitted by the tsetse fly. While the bloodprotein alleles seem to offer some protection against the parasite, they are also associated with kidney disease.

Evolutionary Adaptation Affecting Human Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Skin Color Human skin color is a complex trait controlled by a combination of genes, environmental factors, and their interactions. The skin's outermost layer contains melanocytes, specialized cells containing subcellular organelles called melanosomes that produce and store melanin, the principal light-absorbing pigment. All mammals produce two types of melanin—eumelanin, which is brown–black, and pheomelanin, which is red–yellow. The number of melanosomes, their sizes and shapes, and the relative proportions of the two pigments, as well as other factors including hormones,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

ultraviolet radiation exposure, and aging, produce the primary pigmentation phenotype of human skin. Human skin color is most likely an adaptive response to ultraviolet (UV) radiation. Skin pigment absorbs UV light, which protects the organism from the mutagenic effects of UV described in the Mutation, Repair, and Recombination chapter. UV radiation and human skin color are both correlated with latitude, and dark skin is protective of skin cancer in environments with high UV exposure.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Skin cancer is not the only selective factor for skin color, however. Dark skin is also protective of folate (folic acid), a water-soluble B vitamin essential for DNA synthesis, repair, and gene expression. Folate is especially important in rapidly dividing cells such as those involved in fetal development and spermatogenesis. Folate deficiency results in improper neural tube development, spinal defects, and early fetal loss. Folate is destroyed by UV radiation; however, skin pigmentation shields the molecule. Skin color also plays an important role in regulating the influence of UV radiation on vitamin D, which is synthesized by a UV-sensitive reaction in the skin. Vitamin D is essential for human health because it allows the absorption of calcium from the diet to help build strong bones. Vitamin D deficiency causes rickets, a disease in which cartilaginous precursors of developing bone fail to mineralize. Vitamin D is necessary for normal cell growth and for proper immune system function, and it also inhibits growth of cancer cells. In addition, it is critical for reproduction because rickets in females can result in a narrowed pelvic channel that creates an increased risk of infant and maternal mortality and morbidity. Female infertility and fetal wastage are outcomes of vitamin D deficiency in laboratory mice.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Although vitamin D is available from some foods, diet is only a minor source of vitamin D compared to the amount synthesized in the skin upon UV exposure. Dark melanin pigmentation protects against excess exposure but weakens the absorption of UV light, thereby slowing the process of vitamin D synthesis. For a given amount of UV exposure in a set time period, lighter skin allows the synthesis of more vitamin D compared to darker skin. For this reason, lighter skin color is thought to be selectively advantageous at higher latitudes, where populations are exposed to reduced

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

sunlight and UV radiation. Human skin color, therefore, results from a balance of influences on human health including protection against skin cancer, prevention of folate destruction, and production of sufficient vitamin D. Because skin does not fossilize, there is no definitive record of hominin skin color evolution, but principles of physiology and evolution can guide speculation on how it changed over time. Chimpanzees are lightskinned and covered by body hair. Early hominins, therefore, were probably also light-skinned and had body hair. At some point during human evolution, body hair was lost, perhaps as a way to keep the body cool in response to higher daily activity levels. Paleontologists think that this shift occurred around 1.5–2.0 Ma, when hominins' lower limbs became longer and their daily ranging distances increased. Exposed pale skin poses a problem in areas of high UV radiation because of the increased risk of skin cancer. It is hypothesized that, following the loss of body hair, darker skin was favored in tropical early humans. Later, as humans dispersed out of Africa, selection on skin color changed again as people migrated further from equatorial locales to regions of reduced UV exposure. For some human skin-color genes, evidence suggests that natural selection has changed their frequencies in different populations. For example, the gene SLC24A5 shows very different allele

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

frequencies in Europe versus the rest of the world. An allele with the amino acid replacement Ala111Thr is found at a frequency of 98–100 percent in several European populations, whereas the ancestral allele with Ala111 occurs at a frequency of 93–100 percent in African, East Asian, and native American populations. The SLC24A5 polymorphism explains roughly 25–38 percent of the skin color differences between Europeans and Africans; however, most human genetic polymorphisms do not show such extreme population differences.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Light skin coloration in Asians is influenced, at least in part, by different genetic variants than those responsible for reduced skin pigmentation in Europeans. For example, an allele of the OCA2 gene resulting in His615Arg is present at high frequencies in East Asians but is absent in Europeans and West Africans. This allele results in lower melanin levels in East Asians. Individuals with zero, one, or two copies of the derived 615Arg allele have progressively lighter skin color, and there is evidence for selection of the allele in East Asian populations. Skin color in humans shows convergent evolution. Even within a single human population, convergence of genetic variants for skin color can be extensive. The MC1R gene that controls which type of melanin is produced by melanocytes has more than 30 variants in Europeans that affect its activity. Another MC1R variant, not found in living humans, was present in Neandertals. These different mutations independently lead to melanosomes containing pheomelanin (red–yellow pigment) rather than eumelanin (brown– black pigment) and produce traits such as red and blond hair, freckles, sun sensitivity, and fair skin. This phenotype is likely to have evolved independently many times during the course of human evolution. Hence, similarity in skin color does not necessarily indicate genetic relatedness among individuals.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

SUMMING UP ■ The copy number of the amylase gene in humans correlates with levels of starch in the diet. ■ Mutations in multiple genes that confer resistance to malaria have been identified. ■ The evolution of skin color has involved multiple adaptations. ■ Differences in SLC24A5 genotypes explain at least 25 percent of the skin pigmentation differences between Europeans and Africans.

CHAPTER SUMMARY ■ The ancestral history of protein and nucleic acid molecules can be inferred because the sequences accumulate changes through evolutionary time.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ Phylogenetic analysis of patterns of Alu sequences demonstrate that chimpanzees are the living species that are most closely related to humans. ■ Multiple species of early hominins evolved in Africa, and some later hominins belonging to the genus Homo migrated out of Africa into other parts of the world, beginning 1.8 million years ago. ■ Anatomically modern humans evolved in Africa about 200,000 years ago, and a subgroup migrated to the Middle East approximately 60,000 years ago. ■ Anatomically modern humans coexisted at the same time with other later hominins. DNA sequencing indicates that a small proportion of the genomes of non-African humans (2–3 percent) derives from interbreeding with Neandertals, which

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

probably occurred in the Middle East just after humans first left Africa. ■ Living Melanesians have an additional contribution to their genomes (2–3 percent) from Denisovans, a second archaic hominin species. ■ Despite limited interbreeding with Neandertals and Denisovans, modern humans most likely displaced older hominin populations that they encountered while populating the globe. ■ Approximately 85 percent of the genetic diversity in human populations exists within any local population. Genetic differences between groups are relatively small. ■ The human genome shows many examples of evolutionary adaptation affecting traits such as increased resistance to infectious disease, enhanced digestion of dietary starch, and alterations in skin color.

REVIEW THE BASICS

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

■ What is a gene tree? A species tree? Must they always agree? Explain why or why not. ■ Distinguish between orthologous genes and paralogous genes. Which type of duplication provides the raw material for the evolution of new gene functions? ■ In a DNA-hybridization experiment, what is the melting temperature, and what does the melting temperature tell you about the similarity between the DNA strands? ■ Among a set of gene trees of independent genes in a group of species, why would you expect the majority of gene trees to have the same structure as the actual species tree? ■ What does it mean to say that FOXP2 is a gene related to language but not “the language gene”?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

■ What is the principle of evolutionary parsimony? How is it used to reconstruct features of the common ancestor of humans and chimpanzees? Explain. ■ Would you regard Australopithecus as an example of an early hominin, a transitional hominin, a pre-modern Homo, or an anatomically modern human, and why? ■ Was there ever a time that anatomically modern humans (Homo sapiens) coexisted with both Homo neanderthalensis and Homo erectus? ■ What is the multiregional model for the origin of modern humans? Do genomic sequences support this model? ■ Do you think the term “mitochondrial Eve” is appropriate to describe the coalescence of human mitochondrial DNA sequences? Why or why not? ■ What is serial colonization, and how does this process affect the geographical pattern of genetic diversity? ■ What is admixture? What role has it played in recent human evolutionary history?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

GUIDE TO PROBLEM SOLVING PROBLEM 1 The accompanying illustration is a much simplified tree of human mitochondrial DNA. The dates of modern human origins and the most recent out-of-Africa expansion are based on the mitochondrial molecular clock, calibrated on the assumption that the time of the most recent common ancestor of humans and chimpanzees is 5 million years ago. What would these estimates become if, instead, the human–chimpanzee divergence time were assumed to be 6.5 million years? Would the margin of error estimates also change?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

ANSWER The estimates themselves, as well as the margins of error, should be increased in the ratio 6.5 : 5, or by a factor of 6.5/5 = 1.3. The estimate 170,000 ± 50,000 years, therefore, becomes 220,000 ± 65,000 years, and the estimate 50,000 ± 25,000 years becomes 65,000 ± 32,500 years. PROBLEM 2 The accompanying table shows values of Ka and Ks in the protein-coding region of each of seven genes A–G. Ka is the number of nonsynonymous (amino acid–changing) substitutions per nonsynonymous nucleotide site, and Ks is the number of synonymous substitutions per synonymous nucleotide site. For which gene is the amino acid sequence most highly conserved? Do any of the genes show evidence for positive selection of amino acid changes? ANSWER With data on Ka and Ks, the pattern of evolutionary

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

change in protein-coding regions is best evaluated according to the ratio Ka/Ks. Values of Ka/Ks < 1.0 represent a conserved amino acid sequence (and the smaller the ratio, the stronger the conservation), Ka/Ks = 1.0 is expected with selective neutrality, and Ka/Ks > 1.0 is expected for positive selection of amino acid changes. For these genes, the values of Ka/Ks are 0.015 (gene A), 0.018 (B), 0.094 (C), 0.025 (D), 1.16 (E), 0.062 (F), and 0.027 (G). The smallest value is 0.015; hence, gene A is the most strongly conserved, though only slightly more so than gene B. The only gene for which Ka/Ks > 1.0 is gene E, for which the ratio is

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

1.16. This value would be consistent with positive selection acting on the coding sequence of gene E.

ANALYSIS AND APPLICATIONS 19.1 What is the UPGMA distance matrix for the tree shown here?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

19.2 In the distance matrix shown here, which pair of taxa should be joined first, and what is the resulting UPGMA tree?

19.3 Shown here is the species tree of four species of Old World monkeys designated A–D. If DNA

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

hybridization were carried out with nonrepetitive DNA from these species, for which species pair (or pairs) would the melting temperature be the lowest? For which species pair (or pairs) would the melting temperature be the highest?

19.4 In a DNA hybridization experiment, what relation would you predict among the melting temperatures of nonrepetitive DNA from modern humans (H), Neandertals (N), and the common chimpanzee Pan troglodytes (C). Briefly explain your answer. In the following list, the symbol Tm(AB) represents the melting temperature between nonrepetitive genomic DNA from species A and species B. (a) Tm(HN) < Tm(HC) < Tm(NC) (b) Tm(HN) < Tm(NC) < Tm(HC)

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(c) Tm(HC) < Tm(HN) < Tm(NC) (d) Tm(HC) < Tm(NC) < Tm(HN) (e) Tm(NC) < Tm(HC) < Tm(HN)

19.5 In a DNA hybridization experiment, what relation would you predict among the melting temperatures of nonrepetitive DNA from modern humans (H), the common chimpanzee Pan troglodytes (C), and the bonobo or pygmy chimpanzee Pan paniscus (B). Briefly explain your answer. In the following list, the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

symbol Tm(AB) represents the melting temperature between nonrepetitive genomic DNA from species A and species B. The symbol “≈” means “is approximately equal to.” (a) Tm(HC) ≈ Tm(HB) < Tm(BC) (b) Tm(HC) ≈ Tm(HB) > Tm(BC) (c) Tm(BC) ≈ Tm(HB) < Tm(HC) (d) Tm(BC) ≈ Tm(HB) > Tm(HC)

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

19.6 The accompanying diagram depicts the species tree of gorilla (Gorilla gorilla), common chimpanzee (Pan troglodytes), and human (Homo sapiens). How do these species correspond to the labels A, B, and C?

19.7 The accompanying diagram is a species tree of three species in which the numbered line segments correspond to different branches of the species tree. If a new mutation (yielding a derived allele) occurs and is fixed within an interval corresponding to one of the line segments: (a) In which segment would the mutation yield a derived allele shared by all three species? (b) In which segment would the mutation yield a derived allele shared by species B and C?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

19.8 If the evolutionary split between humans and chimpanzees took place 6.5 million years ago, what is the total evolutionary time separating humans and chimpanzees?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

19.9 A large population of primates undergoing random mating has a copy-number polymorphism consisting of a tandem duplication of a protein-coding gene. In this population, 20 percent of the chromosomes carry one copy of the gene, and 80 percent carry two copies of the gene. What are the expected proportions of individuals with 2, 3, and 4 copies of the gene? 19.10 A large population of primates undergoing random mating has a copy-number polymorphism consisting of tandem copies of a protein-coding gene. In this population, 20 percent of the chromosomes carry one copy of the gene, 30 percent carry two copies of the gene, and 50 percent carry three copies of the gene. What is the expected proportion of individuals with five or more copies? 19.11 In humans, chimps, and gorillas, the gene coding for the blue, light-sensitive, protein pigment opsin is autosomal, whereas the red and green opsins are Xlinked. The situation is more complicated in some New World monkeys, including marmosets, tamarins, and spider monkeys. These species have only one

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

opsin gene in the X chromosome, but it is polymorphic. Some X chromosomes carry a green opsin, whereas others carry a red opsin. Sex determination in these monkeys is similar to sex determination in humans. Consider such a species in which the frequency of the red opsin allele is 0.36 and that of the green opsin allele is 0.64. Assuming that mating is random with respect to this gene, what is the expected frequency of females that have the ability to perceive: (a) Red but not green? (b) Green but not red? (c) Both red and green?

19.12 For the situation described in Question 19.11, what is the expected frequency of males that have the ability to perceive: (a) Red but not green? (b) Green but not red?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(c) Both red and green?

19.13 If the probability that a nucleotide pair in human DNA differs from its counterpart in chimpanzee DNA is 1 percent, what is the probability that, by chance, a region of 1 kb has an identical nucleotide sequence in both species? (Note: Assume that adjacent nucleotide sites are independent.) 19.14 Which of the following conditions make it less likely that a gene tree will correspond to the species tree? (a) A short time between speciation events

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

(b) A large ancestral population size (c) A polymorphism in the common ancestor (d) All of these (e) None of these

19.15 In the Neandertal genome, derived SNPs (not matching those in chimpanzee) have been identified. Those derived SNPs in Neandertals are compared with SNPs found in three modern human populations shown in the accompanying tree.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(a) If more SNPs in human populations 2 and 3 match Neandertal SNPs than do SNPs in human population 1, so that the degree of matching is expressed as H2 = H3 > H1, what does that suggest about hybridization between Neandertals and modern humans? (b) Alternatively, suppose the matching pattern with Neandertal derived SNPs is H3 > H2 > H1. What does that imply about admixture?

19.16 Which model of modern human origins would predict the closest similarity among currently existing human populations? Briefly explain your answer. (a) The multiregional model (b) The African replacement model (c) The hybridization model

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

(d) The models are not distinguishable in this regard.

19.17 Classify the following species according to the likelihood that each may have contributed genes to the genome of modern humans (Homo sapiens) by hybridization. Choices are “not possible,” “highly likely,” and “possibly.” (a) Australopithecus afarensis (b) Homo habilis (c) Homo neanderthalensis (d) Homo erectus

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

19.18 Interpret the following table of Ka and Ks values for a gene present in both humans and gorillas.

19.19 A 900-kb inversion in human chromosome 17 that is rare in Africans has a frequency of about 0.20 in European populations. The inversion haplotype (that is, the particular combination of alleles present in the inversion) is identical from one inverted chromosome to another. Why would the inversion haplotype be expected to be maintained? 19.20 Imagine a population in which a gene undergoes a selective sweep. Explain whether each of the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

following conditions would increase or decrease the size of the haplotype block associated with the favored gene. (a) A larger selection coefficient (b) An increase in the frequency of recombination (c) An increase in the rate of mutation

19.21 A selective sweep of a new, favored mutant allele results in a haplotype block (a region of linked alleles around the mutant) that increases in frequency along with the mutant. How would you expect the size of the haplotype block to depend on the frequency of recombination in the region? How would you expect the size of the haplotype block to depend on the strength of selection?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

19.22 Many of the genetic adaptations that protect against malaria are also associated with anemia due to defective or reduced number of red blood cells. Can you suggest an explanation? 19.23 An organism is polymorphic for two kinds of mutations that affect the reproductive period. One allows reproduction a year earlier, but reproduction terminates at the normal time. The other allows reproduction a year longer, but reproduction begins at the normal time. Which of these mutants (if either) would you expect to be favored by natural selection? 19.24 A species is subject to epidemics of two different infectious diseases. One kills mainly young individuals, the other mainly old individuals. For which infectious disease would you expect natural selection

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

to favor genetic adaptations for resistance more strongly, and why? 19.25 In the serial colonization model, which populations— ancestral or descendant—are expected to have the greater amount of genetic diversity, and why? 19.26 Which of the following are functions of the pigments in human skin? (a) To protect against the mutagenic effects of ultraviolet radiation (b) To promote vitamin D synthesis (c) To protect against the breakdown of folic acid, a B vitamin (d) All of the above

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

19.27 In the strictest definition of the term “species,” a member of one species cannot produce viable and fertile offspring with a member of a different species. Does this strict definition have a bearing on whether Neandertals should be called the species Homo neanderthalensis or the subspecies Homo sapiens neanderthalensis? 19.28 Alleles of a gene are typed from three populations X, Y, and Z. Two types of alleles can be distinguished, A and A'. Each population is fixed for one of the alleles. Which of the following findings are consistent with the gene tree shown here?

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

(a) X fixed for A, Y and Z fixed for A′ (b) X and Y fixed for A, Z fixed for A′ (c) X and Z fixed for A, Y fixed for A′ (d) X and Y and Z fixed for A′

19.29 Would you expect to find lactase persistence in Neandertals? Extra copies of the salivary amylase gene? Why or why not? 19.30 Would you regard hemoglobin S in Africa and hemoglobin H in India as an example of convergent evolution? Why or why not? 19.31 Why are regions of the genome with low or absent recombination more useful for tracing the genetic ancestry of populations? 19.32 In the course of human evolution, has sexual dimorphism increased, decreased, or stayed about the same?

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 Species X, Y, and Z are related to ancestral species W according to the species tree shown here. The population in the branch labeled 1 is polymorphic for alleles A and A′. The alleles A and A′ have equal frequencies, and A′ is the derived allele. Branch 2 is so short that fixation of either allele in this branch has a negligible probability. Assume that one or the other allele becomes fixed independently in each of the branches 3, 4, and 5, in a process called lineage sorting.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

What is the probability that the alleles undergo lineage sorting in such a way that: (a) There is no gene tree (that is, the alleles in X, Y, and Z are identical)? (b) The gene tree is consistent with the species tree? (c) The gene tree is not consistent with the species tree? CHALLENGE PROBLEM 2 In Challenge Problem 1, suppose that branch 2 is long enough that fixation of either A or A′ takes place in branch 2 with a probability of 2/3 and that fixation occurs independently in branch 3. With probability 1/3, fixation of either A or A′ takes place independently in branches 3, 4, and 5. What is the probability that the alleles undergo lineage sorting in such a way that: (a) There is no gene tree (that is, the alleles in X, Y, and Z are identical)? (b) The gene tree is consistent with the species tree? Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(c) The gene tree is not consistent with the species tree?

FOR FURTHER READING Li, J. Z., Absher, D. M., Tang, H., Southwich, A. M., Casto, A., Ramachandran, S., … Myers, R. M. (2008). Worldwide human relationships inferred from genome-wide patterns of variation. Science,

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

319(5866), 1100–1104. http://doi.org/10.1126/science.1153717 A seminal paper, in which SNP variation was used to determine distribution of genetic variation within and between human populations and relationships among populations. Nielsen, R., Akey, J. M., Jakobsson, M., Pritchard, J. K., Tishkoff, S., & Willerslev, E. (2017). Tracing the peopling of the world through genomics. Nature, 541(7637), 302–310. http://doi.org/10.1038/nature21347 A current perspective on the recent evolutionary history of Homo sapiens.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Orlando, L., Gilbert, M. T. P., & Willerslev, E. (2015). Reconstructing ancient genomes and epigenomes. Nature Reviews Genetics, 16(7), 395–408. http://doi.org/10.1038/nrg3935 An excellent overview of the theory and practice of ancient DNA analysis. Salem, A.-H., Ray, D. A., Xing, J., Callinan, P. A., Myers, J. S., Hedges, D. J., … Batzer, M. A. (2003). Alu elements and hominid phylogenetics. Proceedings of the National Academy of Sciences of the United States of America, 100(22), 12787– 12791. http://doi.org/10.1073/pnas.2133766100

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Demonstrates how use of Alu insertion points as phylogenetic characters unambiguously resolves hominid phylogenetic relations. Sarich, V. M., & Wilson, A. C. (1967). Immunological time scale for hominid evolution. Science, 158(3805), 1200–1203. Retrieved from http://www.ncbi.nlm.nih.gov/pubmed/4964406 An important application of molecular clock‒based reasoning to human evolution. In this case, immunological distance was used as a measure of protein divergence, and it resulted in the hypothesis that humans and chimpanzees diverged about 5 million years ago. This figure was controversial at the time, but subsequent work has supported the general time frame proposed.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Wall, J. D., & Yoshihara Caldeira Brandt, D. (2016). Archaic admixture in human history. Current Opinion in Genetics & Development, 41, 93–97. http://doi.org/10.1016/j.gde.2016.07.002 Recent review of genetic evidence for hybridization of Homo sapiens with Neandertals and Denisovans.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-11 22:07:14.

Appendix A: Answers to EvenNumbered Problems Chapter 1 1.2 In regard to replication, they inferred that each strand of the double helix was used as a template for the formation of a new daughter strand having a complementary sequence of bases. In regard to coding capability, they noted that genetic information could be coded by the sequence of bases along the DNA molecule, analogous to letters of the alphabet printed on a strip of paper. Finally, in regard to mutation, they noted that changes in genetic information could result from errors in replication, and the altered nucleotide sequence could be then perpetuated. 1.4 A mixture of heat-killed S cells and living R cells was able to cause pneumonia in mice, but neither heat-killed S cells alone nor living R cells alone could do so.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1.6 Phenol would not destroy the transforming activity, strong alkali would. 1.8 Because the mature T2 phage contains only DNA and protein; the labeled RNA was left behind in the material released by the burst cells. 1.10 In this bacteriophage DNA, the amount of A does not equal that of T, and the amount of G does not equal that of C. You can therefore deduce that the DNA molecule is single stranded, not double stranded. 1.12 5′-CTGAT-3′

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

1.14 The complementary strand has the sequence 5′TACTACTAC…-3′. 1.16 5′-AUCAG-3′ 1.18 The complementary sequence is 5′-AUACGAUA-3′. 1.20 The codon for leucine must be 5′-UUA-3′. 1.22 Six possible reading frames would need to be examined. Either DNA strand could be transcribed, and each transcript could be translated in any one of three reading frames. 1.24 The result means that an mRNA is translated in nonoverlapping groups of three nucleotides: the genetic code is a triplet code. 1.26 The codon 5′-UGG-3′ codes for Trp, and in this random polymer the Trp codon is expected with a frequency of 1/4 × 3/4 × 3/4 = 9/64. The amino acid Val could be specified by either 5′-GUU-3′ or 5′-GUG-3′; the former has an expected frequency of 3/4 × 1/4 × 1/4 = 3/64 and the latter of 3/4 × 1/4 × 3/4 = 9/64, totaling 12/64 or 3/16. The amino acid Phe could be specified only by 5′-UUU-3′ in this random polymer, and so Phe would have an expected frequency of (1/4)3 = 1/64.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

1.28

1.30 a. The amino acid sequence is Met–Val–His–Leu–Thr–Pro– Glu–Glu–Lys–Ser. b. The resulting amino acid sequence would be Met–Val–His–Leu–Thr–Pro–Arg–Arg–Ser–Leu…; all codons following the deletion are changed because of the shift in reading frame by – 1 relative to the nonmutant RNA. c. The resulting amino acid sequence is Met–Val–His–Leu– Thr–Pro. In this case, translation terminates after Pro

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

because the frameshift mutation creates the termination codon 5′-UGA-3′.

Chapter 2 2.2 (e) is the 5′ carbon and carries the phosphate group; (c) is the 3′ carbon and carries the hydroxyl group 2.4 A DNA palindrome reads the same forward and backward, but along opposite strands. a. This is a palindrome because the partner strand reads 3′-GGCC-5′; b. No, because the partner strand reads 3′-AAAA-5′; c. Yes, because the partner strand reads 3′-CGATCG-5′, d. No, because the partner strand reads 3′-GGCGAG-5′; e. No, because the partner strand reads 3′-TTCCAA-5′. 2.6 a. ScaI produces blunt ends; b. NheI produces a 5′ overhang; c. CfoI produces a 3′ overhang.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

2.8

2.10 a. B5 = T; b. B6 = G; c. B7 = the complementary pyrimidine; d. B8 = T or A. 2.12 The probability of a restriction site is 1/4n, where n is the number of nucleotides in the site. a. 1/44 = 0.0039; b. 1/46 = 0.00024; c. 1/48 = 0.000015. 2.14 a. 4.6 × 106/256 ≈ 1.8 × 104; b. 4.6 × 106/4096 ≈ 1.1 × 103; c. 4.6 × 106/65536 ≈ 70.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

2.16 Probe A yields bands of 4 and 12 kb, probe B yields bands of 8 and 12 kb.

2.18 The sites complementary to the primers are shown in bold.

Hence, the amplified fragment has the sequence

2.20 a. 3 kb/3 × 106 kb = 0.0001%; b. (3 × 210 kb)/(3 × 210 kb + 3 × 106 kb) = 0.1%; c. (3 × 220 kb)/(3 × 220 kb + 3 × 106 kb)

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

= 51.2%; d. (3 × 230 kb)/(3 × 230 kb + 3 × 106 kb) = 99.9%. 2.22 The third SNP (C/T) would assist, since all of the fragments that carry the “C” SNP allele also carry the disease allele “D”. The other three SNP's would not, since the SNP and disease alleles are not strictly associated with one another. For example, A fragment with the “A” allele of the first (A/T) SNP could carry either “D” (the top five fragments) or “d” (the bottom fragment). 2.24 Since the evidence is based on total genomic DNA from an individual, all bands found in the evidence must match bands in the perpetrator. Anyone with one or more bands that do not match those in the evidence can be excluded. To find the matching bands in the suspects, for each locus draw straight horizontal lines through the bands of X. For locus 1, suspects

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

C, D, E, F, and G are excluded but A and B are not excluded; for locus 2, suspects B, C, D, E, and G are excluded but A and F are not excluded; and for locus 3, suspects C, D, E, and G are excluded but A, B, and F are not excluded. Taking all three probes together, all suspects can be excluded except for suspect A. This does not necessarily imply that A is the perpetrator, only that A cannot be excluded. Nevertheless, a match of three highly polymorphic DNA markers is highly suggestive. 2.26 After cleavage, phosphate 2 goes with B1 and phosphate 4 goes with B3, so the B1 and B3 mononucleotides will be radioactive.

Chapter 3 3.2 3/4 3.4 3 × 2 × 2 × 1 × 3 = 36 3.6 2/3

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

3.8 Evidently the 7 kb band is a marker for the recessive allele. a. The expected phenotype of II-1 is a single band of 7 kb. b. Because II-2 is not affected, the possible molecular phenotypes are a single band of 3 kb or two bands of 3 kb and 7 kb. 3.10 a. Because the trait is rare, it is reasonable to assume that the affected father is heterozygous HD/hd, where hd represents the normal allele. Half of his gametes contain the HD allele, so the probability is 1/2 that his son received the allele and will later develop the disorder. b. We do not know whether the young man is heterozygous HD/hd, but the probability is 1/2 that he is; if he is heterozygous, half of his gametes will carry the Hd allele. Therefore, the overall probability that his child has the HD allele is (1/2) × (1/2) = (1/4).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

3.12 a. The trait is most likely to be due to a recessive allele because there is consanguinity (mating between relatives) in the pedigree. b. The double line indicates consanguineous mating. c. III-1 and III-2 are first cousins. d. I-1, I-2, II-2 II-3, III-1, and III-2 are most likely Aa, II-1 and II-4 are most likely AA. 3.14 The probability that the parent has the AA BB genotype is Pr{AA BB} = 1/2. The probability of producing an A b gamete, given that a parent has the AA BB genotype, is Pr{A b|AA BB} = 0. The probability that the parent has the AA Bb genotype is Pr{AA Bb} = 1/2. The probability of producing an A b gamete, given that the parent has the AA Bb genotype, is Pr{A b|AA Bb} = 1/2. Therefore, the overall probability of producing an A b gamete is Pr{A b}= (1/2)(0) + (1/2)(1/2) = 1/4. By the same reasoning, the probability of producing an A B gamete is Pr{A B} = (1/2)(1) + (1/2)(1/2) = 3/4. 3.16 (0.002)(1) + (1 – 0.002)(0.002) = 0.004 3.18 Parents have genotypes WW and ww. All F1 progeny from this cross must be heterozygous Ww. The ratio of different phenotypes in the F2 generation is 3/4 round seeds

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

(WW × Ww) and 1/4 wrinkled (ww). Among the F2 round seeds, 2/3 are Ww and 1/3 are WW. Therefore, among the progeny of the cross en masse of (2/3 Ww × 1/3 WW) × ww, the expected proportion with wrinkled seeds will be (2/3) × (1/2) × 1 = 1/3. 3.20 The 9 : 3 : 3 : 1 ratio is that of the genotypes A– B– A– bb, aa B–, and aa bb. The modified 9 : 7 implies that all of the last three genotypes have the same phenotype. The F1 testcross is between the genotypes Aa Bb and aa bb, and the progeny are expected in the proportions 1/4 Aa Bb, 1/4 Aa bb, 1/4 aa Bb, and 1/4 a abb. The last three genotypes would

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

again have the same phenotype, and so the ratio of phenotypes among the progeny of the testcross is 1 : 3. 3.22 Since the woman is affected with the trait, she has a probability of 1/2 of passing the dominant allele to her child, and the child will have a probability of 1/2 of actually exhibiting the trait (because the trait shows 50% penetrance). Thus, the probability that the child will be affected is therefore 1/2 × 1/2 = 1/4. 3.24 Use Bayes theorem with event A as the event that the F1 individual with the dominant phenotype has the genotype Nn (probability 2/3). Event A′ is that the individual is homozygous dominant NN (probability 1/3). Let B be the event that the four offspring from the testcross have the dominant phenotype. Then

3.26 To avoid ambiguity, we will list the bands in the order to which the alleles are written rather than in order of their size. The F1 progeny (A1A2 B1B2) each exhibit four bands of 4, 8, 2, and 6 kb. The F2 progeny consist of 1/16 A1A1 B1B1 (bands of 4 and 6 kb), 2/16 A1A1 B1B2 (bands of 4, 6, and 2 kb), 1/16 A1A1 B2B2 (bands of 4 and 2 kb), 2/16 A1A2 B1B1 (bands of 4, Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

8, and 6 kb), 4/16 A1A2 B1B2 (bands of 4, 8, 6, and 2 kb), 2/16 A1A2 B2B2 (bands of 4, 8, and 2 kb), 1/16 A2A2 B1B1 (bands of 8 and 6 kb), 2/16 A2A2 B1B2 (bands of 8, 6, and 2 kb), and 1/16 A2A2 B2B2 (bands of 8 and 2 kb). 3.28 Paternal/Maternal

1/2 A

1/2 a

1/3 A

1/6 AA

1/6 Aa

2/3 a

2/6 Aa

2/6 aa

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

The expected ratio is 1/6 AA : 3/6 Aa : 2/6 aa.

Chapter 4 4.2 Illustration (A) is anaphase II of meiosis, because the chromosome number has been reduced by half. Illustration (B) is anaphase of mitosis, because the homologous chromosomes are not paired. Illustration (C) is anaphase I of meiosis, because the homologous chromosomes are paired.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

4.4 Since the two genes are in different chromosomes (the mutation for phenylkentonuria is autosomal, that for hemophilia X linked), they will assort independently, and so we can consider them separately. Because both parents are heterozygous for an autosomal recessive phenylketonuria mutation, there is a 1/4 chance that their child will be homozyous recessive. For X-linked hemophilia, only a boy could be affected; a girl would have inherited a normal X chromosome from her father, so she could not have hemophilia. So the probability of a girl being affected with both diseases is 0 and the probability of a boy being affected with both diseases is 1/4 × 1/2 = 1/8. Overall, the probability is (1/2)(0) + (1/2)(1/8) = 1/16. 4.6 Because the child has two Y chromosomes, the nondisjunction took place in the father, during the second meiotic division. 4.8 a. For each gene, half of the offspring will be heterozygous. Since the three genes undergo independent assortment, the proportions for each can be multiplied to get the expected frequency of offspring heterozygous for all three genes: (1/2)3 = 1/8. b. For each gene, half of the offspring will be homozygous, so you can multiply the proportions for each gene to get the expected frequency of offspring homozygous for all three genes: (1/2)3 = 1/8. (Note that the problem does

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

not say that all three genes must be recessive or dominant. So AABBCC, aabbcc, AAbbcc, AAbbCC, etc. are all equivalent.) 4.10 a. (1/2)6 =1/64. b. 6(1/2)6 = 6/64. c. 15(1/2)6 = 15/64. d. 20(1/2)6 = 20/64. e. 1 – (1/64) – (6/64) – (15/64) = 42/64. 4.12 The observed F2 ratio suggests that two different genes are involved in the determination of bulb color. Moreover, since the ratio departs from the classic Mendelian ratio of 9 : 3 : 3 : 1, it means that the genes interact with each other. (They show epistasis.) Let us use the symbol R to denote a dominant allele necessary to develop the red color, and r be a recessive allele of the same gene, so that onion bulbs with the genotype rr are white. Let us also use the symbol W to denote a dominant allele of another gene that suppresses the expression of R. On this hypothesis, bulbs with the genotypes R– W– are also white. Based on this hypothesis, the initial cross was between an RR ww red plant and an rr WW white plant. The F1 progeny of this cross are genotypically Rr Ww and white. The F2

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

progeny still have an underlying genotypic ratio of 9 R– W– : 3 R– ww : 3 rr W– : 1 rr ww. If onion bulbs with the genotype rr ww are yellow, those with the genotype R– ww are red, and those with the genotypes R–W– and rr W– are white, then we can explain the observations because 9 (R– W–) +3 (rr W–) = 12. 4.14 The chi-square value is 4.00. 4.16 There are 21 possible genotypes for an autosomal gene: 6 homozygous and 15 heterozygous. For an X-linked gene the number of possible genotypes is 27: 6 homozygous and 15 heterozygous genotypes in females, and 6 hemizygous genotypes in males. 4.18 a. The probability that their first child will have brown eyes is 3/4 because both parents must be heterozygous. b. 2/3. c. The probability that this couple will have three children with blue 3

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

eyes is (1/4)3 = 1/64. The probability that none of the three children will have blue eyes is (3/4)3 = 27/64. 4.20 For convenience we will use the symbols W and w to denote the wildtype and mutant alleles, respectively, of the Xlinked gene. Let event A be that III-1 is heterozygous Ww. Let event B be that none of IV-1, 2, 3 are affected. The complement of A, the event A′, is the event that III-1 is homozygous WW. Then, Pr{A} = 1/4 and Pr{A′} = 3/4, since these do not depend on B. Furthermore, Pr{B|A′} = 1 and Pr{B|A} = (1/2)2. The reason the 1/2 is squared and not cubed

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

is that the female IV-2 is not informative with regard to III-1; in particular, because the male III-2 has genotype WY, the female IV-2 will be unaffected no matter what the genotype of III-1. With these expressions in place we can use Bayes theorem as follows:

4.22 For convenience we will use the symbols W and w to denote the wildtype and mutant alleles, respectively, of the Xlinked gene. Let event A be that III-1 is heterozygous Ww and let event B be that the individuals IV-1, 2, 3 are all not affected. Since II-2 must be heterozygous Ww, then the probability that III-1 is heterozygous is Pr{A} = 1/2. The complement of A is the event A′ that III-1 is homozygous WW, and so Pr{A′} = 1/2. Since the male III-2 is affected, and none of IV-1, IV-2, and IV3 is affected, all of these offspring must have inherited the W allele from III-1. Hence Pr{B|A} = (1/2)3 and Pr{B|A′} = 1. Applying Bayes theorem to this situation we have

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

4.24 X-linked inheritance is suggested by the fact that each male has only one band, which is transmitted to his daughters but not his sons. The deduced genotypes are: I-1 (A1A1), I-2 (A2), II-1 (A1), II-2 (A1A2), II-3 (A1), II-4 (A1), II-5 (A1A2), II-6 (A2), III-1 (A2), II-2 (A1A2), III-3 (A1A2), and III-4 (A1). 4.26 The accompanying Punnet square shows the outcome of the cross. The X- and Y-chromosomes from the male are denoted in boldface, the attached X-chromosomes are yoked by a ^sign.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

The surviving progeny consists of yellow males and wild-type females in equal proportions. Attached-X inheritance differs from the typical situation in that the sons, rather than the daughters, receive their fathers′ X-chromosome. 4.28 The results of the crosses reveal how the three alleles control plumage pattern. If we let Pr be the allele controlling the restricted pattern, then this allele must be dominant to the Pm and Pd alleles responsible for the mallard and dusky patterns, respectively. Also, the Pm allele is dominant to Pd. a. an F1 male from cross 1 has the genotype Pr Pm; an F1 female from cross 2 has the genotype Pm Pd. Half of their progeny would have the restricted and the other half the mallard plumage patterns. b. In this case, 1/2 of the progeny would have the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

restricted plumage pattern, 1/4 would have the mallard plumage pattern, and 1/4 would be dusky. 4.30 a. All affected individuals, except two (individuals 9 and 17), are homozygous for both FMF and VNTR allele number 1. This implies that the FMF mutation and VNTR1 are close together in the same chromosome. b. Affected individuals 9 and 17 are heterozygous for the VNTR locus, which can be explained by recombination between the VNTR locus and the FMF locus in one of the parents.

Chapter 5 5.2 a, b, c, and d are correct; e, f, and g are false. 5.4 If the frequency of recombination is 0.05, the frequency of choosing a nonrecombinant gamete at random is 1 – 0.05 = 0.95. 5.6 1 : 2 : 1 5.8 100 percent. In real organisms, the maximum frequency of recombination is 50 percent.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

5.10 Three: parental ditype, nonparental ditype, and tetratype. 5.12 Since we know the frequencies of recombination (0.10 and 0.15) and also that the coefficient of coincidence equals 0.40, we can write the frequency of double crossovers that should be observed as 0.40 × 0.10 × 0.15 = 0.006. These are of two types, A b C and a B c, and so the expected number among 1000 gametes is 3 of each. In the a–b interval, the expected proportion of single recombinants equals 0.10 – 0.006 = 0.094. (The trick here is to remember to subtract the double crossovers.) There are two types of single crossovers in this interval, A b c and a B C, and among 1000 gametes the expected number of each is 94/2 = 47. Similarly, in the b–c interval, the expected proportion of single recombinants equals 0.15 – 0.006 = 0.144. In this case, the two types are A B c

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

and a B C, and each has an expected number of 144/2 = 72. The remaining gametes, of which there are 1000 – 6 – 94 – 144 = 756, are expected to be divided equally between the nonrecombinant A B C and a b c types, for an expected number of 756/2 = 378 of each. 5.14 a. This is a 3 : 1 ratio, which means that the F1 parent is homozygous for one gene, in this case R, and heterozygous for the other. The genotype of the F1 parental plant is therefore R T / R t; b. This is also a 3 : 1 ratio, but in this case the genotype of the F1 parental plant is r T / r t; c. Here again is a 3 : 1 ratio, and the genotype of the F1 parental plant is R T / r T; d. In this case, there are four classes of progeny, and so we can assume that F1 parent was doubly heterozygous. We know that the genes are linked, which explains why the ratio is different from 9 : 3 : 3 : 1. The observed distribution of phenotypes in the F2 generation can be explain by the F1 plant having a genotype R T / r t. Although it is not obvious from the data, the frequency of recombination is 20%. You can verify this for yourself with a Punnett square.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

5.16 For each pair of genes, classify the tetrads as PD, NPD, and TT, and tabulate the results as follows:

For this pair of genes, PD = NPD, therefore, leu2 and trp1 are unlinked.

For this pair of genes, PD = NPD, therefore, leu2 and met14 are unlinked.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

For this pair of genes, PD = NPD and therefore, trp1 and met14 are also unlinked. However, there are no TT for trp1 and met14, which means that both genes are closely linked to their respective centromeres. Because they are so close to their centomeres, the segregation of trp1 and met14 can be used as genetic markers of centromere segregation, which allows us to

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

map the leu2 gene with respect to its centromere. Because the other two genes do not recombine with their centromeres, all TT for leu2 will result from recombination of leu2 with its centromere. The map distance between leu2 and its centromere is therefore (1/2) × (100/1000) × 100 = 5 cM. 5.18 a. The most frequent classes of the progeny imply that the genotype of the heterozygous parent is F g H / f G h, with the gene order still to be determined. b. Comparison of the parental classes with the double-recombinants (least common) identifies the gene in the middle, which is G in this case. Thus, the correct linear order of the genes is F G H or H G F. c. The map distance between F and G is (74 + 6)/1000 = 0.08. The map distance between G and H is (114 + 6)/1000 = 0.12. Hence the map distance between the middle gene G and its nearest neighbor F is 0.08 = 8% = 8 map units = 8 cM. d. The map distance between the middle gene G and its farthest neighbor H is 0.12 = 12% = 12 map units = 12 cM. e. The expected number of double crossovers is the product of the frequencies of recombination in each region and the total number of progeny: 0.08 × 0.12 × 1000 = 9.6. f. The coincidence c is the observed number of double crossovers over the expected or 6/9.6 = 0.625. The interference is i = (1 – c) = 1 – 0.625 = 0.375.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

5.20 The frequency of dpy-21 unc-34 recombinants is expected to be 0.24/2 = 0.12 among both eggs and sperm, so the expected frequency of dpy-21 unc-34/dpy-21 unc-34 zygotes is 0.12 × 0.12 = 1.44 percent. 5.22 Mutations 1, 2, and 7 form one complementation group, mutation 2 is the only representative of another complementation group, mutations 4 and 6 make up a third complementation group, mutations 5 and 9 form a fourth complementation group, and a fifth complementation group is represented by mutation 8.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

5.24 Consider each gene in relation to first-division or seconddivision segregation. Gene a shows 1766 asci with the firstdivision segregation and 230 + 14 with the second-division segregation. The map distance between a and its centromere is therefore (1/2) × (234/2000) × 100 = 5.85 map units. Gene b shows 1766 + 14 asci with the first-division segregation and 220 with the second-division. The map distance between b and its centromere is, therefore, (1/2) × (220/2000) × 100 = 5.5 map units. For the two genes together, the asci are 1986 PD, 0 NPD, and 14 TT. Because PD NPD, the genes are linked. The map distance between a and b is (1/2) × (14/2000) × 100 = 0.35 map units. Putting all these results together yields the following genetic map. 5.26 a. The percentage of the testcross progeny expected to be phenotypically A r equals 1/2 of the recombinant progeny, which is 0.30/2 = 0.15, or 15%. b. With no interference (i = 0), crossovers in the adjacent regions are independent. The probability of obtaining a nonrecombinant (parental) chromosome is therefore the product of the probability of no recombination between A and R (1 – 0.30 = 0.70) and the probability of no recombination between R and B (1 – 0.10 = 0.90). This would imply that 0.7 × 0.9 = 0.63, or 63% of the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

gametes should be nonrecombinant, and half of these, or 31.5% would be expected to carry the alleles A R B. 5.28 a. The woman II-3 is necessarily heterozygous for the hemophilia mutation because she had to inherit the X chromosome carrying the hemophilia mutation from her father

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

I-2. Therefore, the probability that the woman II-3 is heterozygous for the hemophilia mutation is 1. Note also that the hemophilia mutation must be in the same chromosome as the f allele of the restriction-fragment length polymorphism (RFLP), since the source of the hemophilia mutation, the male I-2, has only one X chromosome and it carries the f allele as well. The genotype of II-3 is therefore f a/ s A. b. The woman III-2 must have inherited the s allele from her father II-4, and so the f allele band must have come from her mother II-3, who as we have seen has the genotype f a/ s A. In order for the woman III-2 to be heterozygous for the hemophilia mutation, she must have inherited nonrecombinant chromosome from her mother. Since the frequency of recombination is 0.30, the probability of not having a recombination event is 1 – 0.30 = 0.70. c. The bands in the gel imply that IV-1 had to inherit the f allele from his mother III-2, because the X chromosome in a male must come from his mother. We have already seen that the probability that the f-bearing chromosome in III-2 also carries the a allele is 0.70. If it does carry the a allele, then the probability that IV-1 inherits the a allele is the probability that his X chromosome is a nonrecombinant, which is 0.70. Altogether, the probability that IV-1 is affected with hemophilia is the probability that no recombination took place in two successive generations, or (0.70)2 = 0.49.

Chapter 6

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

6.2 The genetic consequence of the obligatory crossover is that alleles in the pseudoautosomal region in the X chromosome can be interchanged with their homologous alleles in the pseudoautosomal region in the Y chromosome, yielding a pattern of inheritance in pedigrees that is indistinguishable from that of ordinary autosomal inheritance. 6.4 a. The Klinefelter karyotype is 47, XXY, hence one Barr body; b. the Turner karyotype is 45, X, hence no Barr bodies; c. the Down syndrome karyotype is 47, +21; people with this condition have one or zero Barr bodies depending on whether they are female (XX) or male (XY); d. males with the karyotype 47, XYY have no Barr bodies; e. females with the karyotype 47, XXX have two Barr bodies. 6.6 The only 45-chromosome karyotype found at appreciable frequencies in spontaneous abortions in 45, X, and so 45, X is the probable karyotype. Had the fetus survived, it would have been a 45, X female with Turner syndrome.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

6.8 Because the X chromosome in the 45, X daughter contains the color-blindness allele, the 45, X daughter must have received the X chromosome from her father through a normal X-bearing sperm. The nondisjunction must therefore have occurred in the mother, resulting in an egg cell lacking an X chromosome. 6.10 The inversion has the sequence A B E D C F G, the deletion A B F G. The possible translocated chromosomes are (a) A B C D E T U V and M N O P Q R S F G or (b) A B C D E S R Q P O N M and V U T F G. One of these possibilities includes two monocentric chromosomes and the other includes a dicentric and an acentric. Only the translocation with two monocentrics is genetically stable. 6.12 The mother has a Robertsonian translocation that includes chromosome 21. The child with Down syndrome has 46

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

chromosomes, including two copies of the normal chromosome 21 plus an additional copy attached to another chromosome (the Robertsonian translocation). This differs from the usual situation, in which Down syndrome children have trisomy 21 (karyotype 47, +21). 6.14 The genes uncovered by a single deletion must be contiguous. The gene order is deduced from the overlaps between the deletions. The overlaps are a and d between the

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

first and second deletions and e between the second and third. The gene order (as far as can be determined from these data) is diagrammed in part A of the accompanying illustration. The deletions are shown in red. Gene b is at the far left, then a and d (in unknown order), then c, e, and f. (Gene c must be to the left of e, because otherwise c would be uncovered by deletion 3.) Part B is a completely equivalent map with gene b at the right. The ordering can be completed with a three-point cross between b, a, and d or between a, d, and c or by examining additional deletions. Any deletion that uncovers either a or d (but not both) plus at least one other marker on either side would provide the information to complete the ordering.

6.16 The order is (a e) d f c b, where the parentheses mean that the order of a and e cannot be determined from these data. The answer b c f d (a e) is also correct. 6.18 The group of 4 synapsed chromosomes implies that a reciprocal translocation has taken place. The group of 4

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

includes both parts of the reciprocal translocation and their nontranslocated homologues. 6.20 The observation that the cross between these species yields a hybrid in which the chromosomes do not pair indicates that the chromosomes of the two parental species are quite different genetically. One possibility for the origin of G. hirsutum is that it arose by polyploidization in such a hybrid, which would account for the 26 chromosome pairs in G. hirsutum. The backcross results are also consistent with this hypothesis, because each yields a hybrid with 13 bivalents and 13 univalents. The data therefore suggest that present-day American cultivated cotton originally arose as an allotetraploid from a hybrid between a wild species related to G. arboreum and another wild species related to G. thurberi. 6.22 An AABBBBCC type of polyploid could be formed by 9 possible types of hybridization and endoreduplication: [(AA × BB) × (BB)] × CC)], [(AA × BB) × (CC)] × BB)], [(BB × CC) × (BB)] × AA)], [(BB × CC × (AA)] × BB)], [(BB × BB) × (AA)] × CC)], [(BB × BB) × (CC)] × AA)], [(AA × CC) × (BB)] × BB)], [(AA × BB) × (BB × CC)], [(AA × CC) × (BB × BB)]. 6.24 In this male, the tip of the X chromosome containing y+

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

became attached to the Y chromosome, yielding the genotype y X / y+ Y. The gametes are y X and y+ Y, and so all of the offspring are either yellow females or wild-type males. 6.26 Viable progeny are expected to result only from adjacent segregation, in which the Y chromosme segregates with the normal chromosome 2, and the two parts of the X–2 translocation segregate together. The expected result is 50 percent ruby, dumpy males and 50 percent nonruby, nondumpy females. 6.28 The unusual yeast strain is disomic for chromosome 2, and the result of the cross is a trisomic. Segregation of

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

markers on chromosome 2, including his7, is aberrant, but segregation of markers on the other chromosomes is completely normal.

Chapter 7 7.2 Since the variance equals the square of the standard deviation, the only possibilities are that the variance equals either 1 or 0. 7.4 Broad-sense heritability is the proportion of the phenotypic variance attributable to all differences in genotype, or , which includes dominance effects and interactions between alleles. Narrow-sense heritability is the proportion of the phenotypic variance due only to additive effects, or , which is used to predict the resemblance between parents and offspring in artificial selection. If all allele frequencies equal 1 or 0, both heritabilities equal zero. 7.6 For fruit weight, is given. Because , then . Therefore, 87 percent is the broad-sense heritability of fruit weight. The values of H2 for soluble-solid content and acidity

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

are 91 percent and 89 percent, respectively. 7.8 a. Phenotypes 0, 2, 4, and 6 with frequencies 0.015625, 0.140625, 0.421875, and 0.421875, respectively. b. Mean 4.50, variance 2.25, standard deviation 1.50. c. Phenotypes 0, 1, 2, 3, 4, 5, and 6 with frequencies 0.0156, 0.0938, 0.2344, 0.3125, 0.2344, 0.0938, and 0.0156, respectively. Mean 3.00, variance 1.50, standard deviation 1.22. 7.10 a. To determine the mean, multiply each value for the pounds of milk produced (using the midpoint value) by the number of cows, and divide by 300 (the total number of cows) to obtain a value of the mean of 4080 pounds. The estimated 2

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

variance is 626,100 pounds2, and the estimated standard deviation, which is the square root of the variance, is 791 pounds. b. The range that includes 68 percent of the cows is the mean minus the standard deviation to the mean plus the standard deviation, or 4080 – 791 = 3289 pounds to 4080 + 791 = 5662 pounds. c. For 95 percent of the animals, the range is within two standard deviations, or 2498 – 5698 pounds. 7.12 a. The genotypic variance is 0 because all F1 mice have the same genotype. b. With additive alleles, the mean of the F1 generation will equal the average of the parental strains. 7.14 The phenotypic variance, , in the genetically heterogeneous population is the sum . The variance in the inbred lines equals the environmental variance (because in a genetically homogeneous population); hence . Therefore, in the heterogeneous population equals 40 –10 = 30. The broad-sense heritability H2 equals percent. b. If the inbred lines were crossed, the F1 generation would be genetically uniform, and consequently the predicted variance would be

.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

7.16 The offspring are genetically related as half-siblings, and Table 7.3 says that the theoretical correlation coefficient is h2/4. 7.18 a. Because Difflugia is an asexual organism, parent and offspring have the same genetic relationship as monozygotic twins; hence, the broad-sense heritabilities are 0.956 and 0.287, respectively. b. The environmental component of the variance for length of longest spine is larger than that for tooth number. 7.20 and D = 1.689 – (– 0.137) = 1.826. The minimum number of genes is estimated as n = (1.826)2/(8 × 0.0426) = 9.8. 2

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

7.22 h2 = (2.33 – 2.26)/(2.37 – 2.26) = 63.6 percent 7.24 Use Equation 7-6 with M = 10.8, M* = 5.8, and M = 8.8. The narrow-sense heritability is estimated as h2 = (8.8 – 10.8)/(5.8 – 10.8) = 0.40. 7.26 The mean weaning weight of the 20 individuals is 81 pounds, and the standard deviation is 13.8 pounds. The mean of the upper half of the distribution can be calculated by using the relationship given in the problem, which says that the mean of the upper half of the population is 81 + (0.8)(13.8) = 92 pounds. Using Equation 7-6, we find that the expected weaning weight of the offspring is 81 + 0.2(92 – 81) = 83.2 pounds. 7.28 a. The narrow-sense heritabilities are 1.0, 0.67, 0.18, 0.10, 0.02, 0.01, and 0.002, respectively. b. The magnitude of the narrow-sense heritability depends on the parent–offspring correlation in phenotype. With rare recessive alleles, most homozygous offspring come from matings between heterozygous genotypes, so there is essentially no parent– offspring correlation in phenotype. 7.30 Successive substitutions in this case yield Mt = M0 + h2(S1 + S2 + S3 + … + St–1).

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Chapter 8 8.2 The final dilution should contain 700 viable cells per ml, which is a 106-fold dilution of the original suspension. You could do any combination of 100-fold and 10-fold dilutions that multiply to 106, but the simplest would be three successive 100-fold dilutions. 8.4 The T2h plaques will be clear because both strains are sensitive; the T2 plaques will be turbid because of the resistance of the B/2 cells.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

8.6 The E. coli strain contains 5,000 kilobase pairs, which implies approximately 50 kb per minute. Because the l genome is 50 kb, the genetic length of the prophage is one minute. 8.8 Plate on minimal medium that lacks leucine (selects for Leu+) but contains streptomycin (selects for Str-r). The leu allele is the selected marker, and str-r the counterselected marker. 8.10 a. Use m = 2 – 2d1/3, in which m is map distance in minutes and d cotransduction frequency. With the values of d given, the map distances are 0.74, 0.41, and 0.18 minutes, respectively. b. Use d = [1 – (m/2)]3. With the values of m

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

given, cotransduction frequencies are 0.42, 0.12, and 0, respectively. For markers greater than two minutes apart, the frequency of cotransduction equals zero. (c) Map distance a–b equals 0.66 minutes, b–c equals 1.07 minutes, and these are additive, giving 1.73 minutes for the distance a–c. The predicted cotransduction frequency for this map distance is 0.002. 8.12 a. The identical bands were present in the original strain. b. The new bands are the result of transposition of IS5 to new locations in the genome. c. There are three transpositions per 12 lineages per 50 years, 3/(12 × 50) = 0.005 transpositions per lineage per year. d. Each colony has five elements that were present in the original strain, so the transposition rate per element is 0.005/5 = 0.001 transpositions per element per year in each lineage. 8.14 a is true because many mating pairs will break apart after the entry of a+ but before the entry of c+; b is false because b+ has to enter before c+; c is true because many of the b+ str-r colonies will be a− b+ str-r; d is false because many of the b+ str-r colonies will be a− b+ str-r; e is true because more recombination events are needed to result in a+ b− c+ str-r than +

+

+

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

in a+ b+ c+ str-r; f is true because more recombination events are needed to result in a+ b− c+ str-r than in a− b+ c+ str-r; g is true because many mating pairs will break apart after the entry of a+ but before the entry of b+. 8.16 Apparently, h and tet are closely linked, and so recombinants containing the h+ allele of the Hfr tend to also to contain the tet-s allele of the Hfr, and these recombinants are eliminated by the counterselection for tet-r. 8.18 Depending on the size of the F′ plasmid, it could be F′ g, F ′ g h, F′ g h i, and so forth, or F′ f, F′ f e, F′ f e d, and so forth. 8.20 The order 500 his+, 250 leu+, 50 trp+ implies that the genes are transferred in the order his+ leu+ trp+. The met mutation is the counterselected marker that prevents growth of the Hfr parent. The medium containing histidine selects for leu+ trp+, and the number is small because both genes must be incorporated by recombination. 8.22 The time of 15 minutes is too short to allow the transfer of a terminal gene. Thus, one explanation for a z+ str-r colony is that aberrant excision of the F factor occurred in an Hfr cell, yielding an F′ plasmid carrying terminal markers. When the F′ z+ plasmid is transferred into an F− recipient, the genotype of the resulting cell would be F′ z+/z− str-r. Alternatively, a z+ str-r Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

colony can originate by reverse mutation of z− to z+ in the F− recipient. The possibilities can be distinguished because the F′ z+/z− str-r cell would able to transfer the z+ marker to other cells. 8.24 a. Consider each Hfr in turn and, starting with the earliestentering gene, write down the name of each gene as it enters. The difference in time of entry between adjacent genes is the distance in minutes between the genes. When a partial genetic map of gene transfer from each Hfr has been made, arrange

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

the maps so that any shared markers between two or more Hfr strains coincide. This process yields the genetic maps shown below, where the arrows indicate direction of transfer and the numbers are times of transfer in minutes.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Now arrange the composite genetic map in the manner required, in the form of a circle of 100 minutes, with 0 minutes at the top and leu at about 2 minutes. The map is as shown here.

b. Comparison of this map with that in Figure 8.12 indicates that a group of markers present in a contiguous segment of chromosome is in the opposite order as compared with the standard map. The simplest explanation is that the strain in question has undergone an inversion of part of the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

chromosome, with breakpoints in the tonA–pro and trp–terC intervals in the standard map. 8.26 The accompanying map shows the genetic intervals defined by the endpoints of the deletions described in Problem 8.25 and the locations of the mutations with respect to these intervals. Also shown are the boundaries of the rIIA and rIIB cistrons as defined by the complementation test in Problem 8.27.

8.28 The mean number of P2 phage per cell is 3, and the probability of a cell remaining uninfected by P2 is e−3 = 0.050; the mean number of P4 phage per cell is 2, and the probability of a cell remaining uninfected by P4 is e−2 = 0.135. a. Not be infected with either phage: 0.050 × 0.135 × 108 = 6.75 × 105; Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

b. Survivors: 6.75 × 105 + 0.050 × (1 – 0.135) × 108 = 5.00 × 106. c. Producers of P2 progeny: (1 – 0.050) × 0.135 × 108 = 1.28 × 107. d. Producers of P1 progeny: (1 – 0.050) × (1 – 0.135) ×108 = 8.22 × 107.

Chapter 9 9.2 DNA structure is a long, thin thread, and its total length in most cells greatly exceeds the diameter of a nucleus, even allowing for a large number of fragments per cell. This forces

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

you to conclude that each DNA molecule must be folded back on itself repeatedly. 9.4 Each nucleosome includes 145 bp, and with a linker size of 55 bp, the total DNA length per nucleosome is 200 bp. This value corresponds to 5 nucleosomes per kb and 250 nucleosomes per 50 kb. 9.6 No, because the E. coli chromosome is circular. There are no termini. 9.8 The telomers would become longer in each successive generation. 9.10 In Drosophila, the heterochromatin occupies about 25 percent of each chromosome arm nearest the centromere; these regions are not highly replicated in polytene nuclei but clump together to form the chromocenter. The outer 75% of each chromosome arm consists of euchromatin that is highly replicated and that forms the banded regions.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

9.12 The molecules are of the same length, so the molecule with the greater proportion of G–C base pairs will be more stable, and so will require a higher temperature for denaturation, because each G–C pair has three hydrogen bonds whereas each A–T base pair has only two. Hence the duplex with 8 G–C base pairs will be more stable than the duplex with only 4. 9.14 42,000 meters is 42,000,000 millimeters. Scaling this length back down by a factor of 500,000 yields 84 millimeters. A length of 84 millimeters is about the actual length of the fully extended DNA duplex in human chromosome 2. 9.16 The chromosome becomes successively shorter with each round of DNA replication. The new mutations to w result from the successive shortening of the chromosome, when DNA replication can no longer replicate the sequences needed for wild-type expression of the gene.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

9.18 The junction has the sequence shown here

9.20 a. There are 10 base pairs for every turn of the double helix. With four turns of unwinding, 4 × 10 = 40 base pairs are broken. b. Each twist compensates for the underwinding of one full turn of the helix; hence, there will be four twists. 9.22 A change of 51 in the percentage of G + C base pairs results in a change of 20.9°C in Tm, hence the relationship is Tm = 78 + (20.9/51)(g – 22), or Tm = 69.0 + 0.41g °C. 9.24 Because g = (Tm – 69.0)/0.41, the answers are: a. g = 38.3 percent G + C; b. g = 43.9 percent G + C. 9.26 If you double the concentration, renaturation will take place twice as fast because the complementary strands will find each other more rapidly. Hence the C0t1/2 value will remain the same. This result implies that the estimated complexity of a DNA sample does not depend on the initial concentration, since any change in initial concentration is balanced by a corresponding change in the time of renaturation.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

9.28 a. The helix is overwound one complete turn. b. Supercoiling does take place. c. Positive supercoiling relieves overwinding. 9.30 A schematic C0t curve showing the characteristics described in the problem is shown in the accompanying illustration.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

Chapter 10 10.2 The 3′-to-5′ exonuclease activity of DNA polymerase has the critically important proofreading function. If this errorcorrecting mechanism is incapacitated, the frequency of mutations resulting from the incorporation of wrong nucleotides will increase by a factor of about hundred (see Chapter 14 for more detail). 10.4 This protein is called a helicase.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

10.6 Rolling circle replication begins with a single-strand break, which produces 5′-P and 3′-OH groups. The polymerase extends the 3′-OH terminus and displaces the 5′-P end. Since the DNA strands are antiparallel, the strand complementary to the displaced strand must be terminated by a 3′-OH group. 10.8 c. density-gradient centrifugation 10.10 The 5′ carbon carries the group a, which is –CH2OH. In a nucleotide, a corresponds to one or more phosphate groups; b is –H; c is –OH; d is –H, and e corresponds to one of the bases, A, T, G, or C. 10.12 DNA polymerase requires a single-stranded template, a primer, and all four trinucleotides; it adds nucleotides only to an existing 3′ end. In the molecule on the left, the DNA polymerase

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

could elongate each 3′ end using the other strand as a template. The resulting product would have the sequence

In the molecule on the right, on the other hand, the 3′ ends cannot be elongated, as there is no template. In other words, a 5′ overhang (as in the molecule on the left) can be filled in by DNA polymerase, but a 3′ overhang (as in the molecule on the right) cannot. 10.14 For the semiconservative mode of replication, after one round of replication all DNA molecules would be 15N–14N; after two rounds half would be 15N–14N and the other half 14N–14N; and after three rounds of replication, 1/4 would be 15N–14N and 3/4 would be 14N–14N. For the conservative mode, after one round of replication, half of DNA molecules would be 15N–15N and another half 14N–14N; after two rounds, 1/4 would be 15N–15N and 3/4 14N–14N; and after three rounds, 1/8 15N–15N

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

and 7/8 14N–14N. 10.16 a. In each complete 360 degree turn of the double helix, there are 10 bp of DNA. Hence 800 bp of DNA corresponds to 80 complete turns, so each replication fork makes 80 revolutions per second, and hence 80 × 60 = 4800 revolutions per minute. b. Two replication forks move away from the origin of replication in opposite directions, which meet when each has moved half-way around the genome. The time required is (0.5 × 4.3 × 106)/800 sec = 2687.5 sec. or 44.79 min. 10.18 The percentage of adenine in the DNA would have increased by 5 percent. 10.20 DNA replication could occur nearly normally at 37°C but not at 42°C. At the higher temperature, the mutant would not be able to grow. It would be a lethal.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

10.22 The DNA would not necessarily have the same nearest neighbors. If the species both had random sequences, similar base ratios would indeed result in similar nearest-neighbor frequencies. But consider the following counterexample: One of the species has long stretches of T's alternating with long stretches of G's, whereas the other species has a random sequence. In this case, the base ratios would be the same, but the nearest-neighbor frequencies would be very different. 10.24 Rolling circle replication begins with a single-stranded nick, which produces a 5′-P and 3′-OH group. DNA polymerase adds nucleotides to 3′-OH end using the intact (unnicked) strand as a template; this growing strand is the leading strand. The displaced strand with the free 5′-P is the lagging strand. 10.26 a. The first nucleotide in the DNA that is copied is an A, thus the first base in the RNA is U. RNA grows in the 5'-to-3' direction; therefore, the sequence of the eight-nucleotide primer is 5'-UGCAGGUC-3'. b. The C at the 3' end has the – OH group. c. The U at the 5' end has a triphosphate. d. The replication fork is moving in a right-to-left direction, because this will entail lagging-strand synthesis (discontinuous replication) of the strand shown.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

10.28 All three statements are true: a and b follow from the fact that A pairs with T and G pairs with C. Statement c is another way to express b.

Chapter 11 11.2 a. 2; b. 3; c. 2; d. 1. 11.4 a. T102I results from a C → T transition; b. P106S results from a C → T transition; c. Glyphosate-resistant mutants are so rare because they are double mutants, and the occurrence of two mutations in the same gene is a much rarer event than that of either mutation alone.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

11.6 Such people are somatic mosaics and can be explained by somatic mutations in the pigmented cells of the iris of the eye or in their precursor cells. 11.8 a. The start codons are AUG and AUM. b. The stop codons are UAA, UAG, UAM, UGA, and UMA. 11.10 Any single nucleotide substitution creates a new codon with a random nucleotide sequence, so the probability that it is a chain-termination codon (UAA, UAG, or UGA) is 3/64, or approximately 5 percent. 11.12 The other half encode Met–His–Gly–Gly–Gly–Met–His. These clones originate from ligation of the original DNA fragment in an inverted orientation. 11.14 The Lac phenotype is Lac–. Filling in the two-base overhangs was a mistake, as the ligated fragment will have a two-base insertion, which creates a frameshift mutation.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

11.16 The possible lysine codons are AAA and AAG, and the possible glutamic acid codons are GAA and GAG. Therefore, the mutation results from an AT to GC transition in the first position of the codon. 11.18 Single nucleotide substitutions that can account for the amino acid replacements are (1) Met (AUG) to Leu (UUG), (2) Met (AUG) to Lys (AAG), (3) Leu (CUN) to Pro (CCN), (4) Pro (CCN) to Thr (ACN), and (5) Thr (ACR) to Arg (AGR), where R stands for any purine (A or G) and N for any nucleotide. Substitutions (1), (2), (4), and (5) are transversions, and substitution (3) is a transition. Because 5-bromouracil primarily induces transitions, the Leu → Pro replacement is expected to be the most frequent. 11.20 The 2-aminopurine (Ap) could be incorporated in place of adenine opposite thymine, forming an ApT base pair. In the following rounds of replication, the Ap will usually pair with T, but when it occasionally pairs with C, it forms an ApC pair. In

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

the next round of replication, the C pairs with G forming a mutant GC pair. Hence, in this case, the type of mutation induced by 2-aminopurine is a change from an AT pair to a GC pair, which is a transition mutation. Alternatively, and less likely, the Ap could be incorporated opposite C as an ApC base pair. In subsequent rounds of replication, the Ap will usually pair with T, which in the next round of replication will pair with A, forming an AT base pair. In this mode of mutagenesis, the mutation is GC to AT transition. 11.22 Solve (1 × 10−5) × m × 2700 = 0.01, where m is the proportion of X-linked genes capable of mutating to a recessive lethal. Then m = 37%. 11.24 a. If the mutation rate is 0.003 mutation per genome per DNA replication, then 1 – 0.003 = 99.7% of the genomes contain no new mutations after one round of replication. b. The rate of mutation per base pair per DNA replication λ is 0.003/50,000 = 6 × 10−8. For the other genomes, the values are calculated similarly. They equal 6.5 × 10−10 for E. coli and 2.2 × 10−10 for S. cerevisiae. c. The great variation in

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

spontaneous mutation rate per base pair among these genomes suggests that the larger genomes have evolved lower mutation rates per nucleotide pair through either greater fidelity of DNA replication or greater efficiency of repair mechanisms. 11.26 a. At 1 Gy, the total mutation rate is double the spontaneous mutation rate. b. Because a dose of 1 Gy increases the rate by 100%, then a dose of 0.5 Gy increases it by 50%. c. Because a dose of 1 Gy increases the rate by 100%, then a dose of 0.1 Gy by 10%. 11.28 Deduce the sequence of the reading frame of both of the double mutants. The +G with –C combination has the reading frame AGAGGUAAGTCTAGA, which translates as Arg–Gly– Lys–Ser–Arg. This polypeptide has three adjacent amino acid

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

replacements but could function if the replacements could be tolerated. On the other hand, the –G with +C mutant combination has the reading frame AGATAACGCCTCAGA, which translates as Arg–End, where End represents the UAA stop codon. In this double mutant, the downstream duplication does not correct the upstream termination codon. 11.30 If the Holliday junction is resolved with no recombination of the outside markers, one molecule should be “light” and the other “heavy.” If the Holliday junction is resolved with recombination of the outside markers, one molecule should be “light” on the left-hand side and “heavy” on the right-hand side, or the other way around; in this case, the density of both molecules would, on the average, be intermediate.

Chapter 12

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

12.2 a. All pyrimidines: Phe, Ser, Leu, and Pro. b. All purines: Lys, Arg, Glu, and Gly. 12.4 a. The original DNA must have the sequence of bases 5′ATG-3′/3′-TAC-5′, where the bottom strand (written to the right of the slash) is the one transcribed; the inversion has the sequence 5′-CAT-3′/3′-GTA-5′, where again the bottom strand is transcribed. The mRNA therefore contains 5′-CAU-3′ (the “reverse complement” of the original codon), which specifies His. b. None of the reverse-complement codons are synonymous. This answer can be seen by checking the reverse complement of each codon in turn, or more easily by noting that there are no synonymous codons with a complementary base in the second position. c. The chain-termination codons are 5′-UAA-3′, 5′-UAG-3′, and 5′-UGA-3′, which have reverse complements 5′-UUA-3′ (Leu), 5′-CUA-3′ (Leu), and 5′-UCA-3′ (Ser), respectively. d. Same answer as in part c.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

12.6 Codons are read from the 5′ end of the mRNA. The first amino acid in the growing polypeptide is at amino end. Therefore, if the C were added to the 3′ end, the last codon would be UUC, which also codes for the phenylalanine owing to the degeneracy of the genetic code. On the other hand, adding the C to the 5′ end would make the first codon CUU, which specifies valine (Val) at the amino end of the polyphenylalanine chain. 12.8 Translation could be in one of three reading frames: AAG AAG AAG …., AGA AGA AGA …., or GAA GAA GAA ….. The first codes for polylysine, the second for polyarginine, and the third for polyglutamic acid. 12.10 The RNA would be translated as an alternating sequence of the codons AUA–UAU–AUA–UAU– ...., which corresponds to a polypeptide with the repeating sequence Ile–Tyr–Ile–Tyr .... 12.12 In a random sequence there are 23 = 8 possible codons.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Their identities, coding capacities, and frequencies are

Therefore, the amino acids expected in a random polymer and their frequencies are glycine (4/64 = 6.25%), valine (12/64 = 18.75%), tryptophan (3/64 = 4.69%), leucine (9/64 = 14.06%), cysteine (9/64 = 14.06%), and phenylalanine (27/64 = 42.19%).

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

12.14 a. UGYN would be translated as UGY–GYN, which corresponds to Cys–Val (if Y is a T) or Cys–Ala (if Y is a C). Hence, cystein could be followed only by valine or alanine. b. UUYN would be translated as Phe–Phe, Phe–Leu, or Phe–Ser. c. UGGN would be translated as Trp–Gly. d. AUGN would be translated as Met–Cys, Met–Trp (unless N is A, in which case Met is the final amino acid in the polypeptide chain). 12.16 a. AUGN would be translated as AUG–UGN, so the second amino acid could be Cys or Trp. b. In this case, translation would be as AUG–GNN, so the second amino acid could be Val, Ala, Asp, Glu, or Gly. 12.18 5′-UGG-3′ codes for tryptophan. This sequence can pair with either of the anticodons 3′-ACC-5′ or 3′-ACU-5′ anticodons. However, 3′-ACU-5′ can also pair with 5′-UGA-3′, which is a chain-termination codon. If this anticodon were used for tryptophan, the amino acid would be inserted instead of terminating the chain. Therefore, only 3′-ACC-5′ is used. 12.20 The probability of a correct amino acid at a particular site is 1 – 5 × 10−4 = 0.9995, and so the probability of all 300 amino acids being correctly translated is (0.9995)300 = 0.86, or

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

86%. 12.22 If transcribed from left to right, the transcribed strand is the bottom one, and the mRNA sequence is AGA CUU AGC GCU AAA CGU GGU, which codes for Arg–Leu–Ser–Ala–Lys– Arg–Gly. To invert the molecule, the strands must be interchanged in order to preserve the correct polarity, and so the sequence of the inverted molecule is:

This codes for an mRNA with sequence AGA ACG UUU AGC GCU AAG GGU, which codes for Arg–Thr–Phe– Ser–Ala–Lys– Gly.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

12.24 The missing part of the stem is

The stem and loop structure is shown here.

12.26 With no A–to–I modification, the polypeptide sequence is Val–Pro–Arg–Ser–Ser–His; with A–to–I modification, the mRNA (separated into codons) reads 5′-GUI CCI CGC UCG UCU CIU-3′, which is translated as Val–Pro– Arg–Ser–Ser–Arg.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

12.28 Because either strand could be transcribed, and in the transcript there are three possible reading frames, there are six ways in which the DNA could be transcribed and translated. The mRNA will have the same sequence as the nontranscribed strand (with U replacing T), and so the RNA sequences could be as follows:

The red triplets are stop codons. In this case you are lucky because the mRNA at the top has stop codons in all three reading frames and the bottom mRNA has stop codons in two of the three. The reading frame of the mRNA that could be a coding sequence is

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

in which the complete codons specify Lys–Leu–Gly–His. Hence, the transcribed strand is the top strand and the reading frame is the one specified here. 12.30 a. 5′-UACCAUGUCAAACAGCGUAUGGUAGCA GUG-3′. b. Met–Ser–Asn–Ser–Val–Trp.

Chapter 13 13.2 The enzymes are expressed constitutively because glucose is metabolized in virtually all cells. However, the levels of enzyme are regulated to prevent runaway synthesis or inadequate synthesis. 13.4 The most likely level of regulation is (a) transcription, (b) RNA processing, (c) translation.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

13.6 There are two possibilities. The first is that there are two repressor-binding sites (one contained in the EcoRI–BamHI fragment and the other in the HindIII–SacI fragment), which act synergistically. Each alone allows about a 10-fold repression, but together the repression is 100 fold. This hypothesis could be tested by sequencing the restriction fragments; if each contains a repressor-binding site, a region of similar sequence should be found in each fragment. The other hypothesis is that the repressor-binding site is very large, extending across the entire EcoRI–SacI region. In this case, the BamHI–HindIII fragment should also be part of the repressor-binding site, so deletion of this fragment should have a major effect, whereas, with two separate binding sites, the deletion would have at most a minor effect. 13.8 An inversion not only inverts the sequence, but because the DNA strands in a duplex are antiparallel, it exchanges the strands themselves. The inversion of a promoter sequence would therefore have a profound effect because the inverted promoter is in the opposite strand pointing in the wrong

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

direction. On the other hand, one would expect little or no effect of the inversion of an enhancer element, because enhancers are targets for DNA binding proteins irrespective of their strand or orientation in the DNA. 13.10 At 37°C, both lacZ− and lacY− would be phenotypically Lac−; at 30°C, the lacY− strain would be Lac+ and the lacZ− strains would be Lac−. 13.12 The cAMP concentration is low in the presence of glucose. Both types of mutation are phenotypically Lac−. The cAMP-CRP binding does not interfere with repressor binding because the DNA binding sites of cAMP–CRP and lacI are distinct. 13.14 With a repressor, the inducer is usually an early (often the first) substrate in the pathway, and the repressor is inactivated by combining with the inducer. With an aporepressor, the effector molecule is usually the product of the pathway, and the aporepressor is activated by the binding. 13.16 The attenuator is not a binding site for any protein, and RNA synthesis does not begin at an attenuator. An attenuator is strictly a potential termination site for transcription.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

13.18 a. The histidine operon would initiate transcription when the level of histidine is low enough that the His aporepressor is no longer bound, derepressing transcription. b. Attenuation would take place (halting transcription) when the level of charged tryptophan tRNA is high enough that translation of the leader polypeptide can take place. c. Transcription would continue through to the end of the operon when the level of charged tryptophan tRNA is so low that translation stalls at the Trp codons in the leader sequence. 13.20 Amino acids that normally have among the lowest levels of charged tRNA in the cell should have few codons in the leader mRNA, otherwise stalling would take place even at the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

normal low levels. On the other hand, amino acids that have among the highest levels of charged tRNA in the cell should include multiple codons in the leader mRNA, otherwise stalling might not take place often enough. 13.22 The a1 protein functions only in combination with the α2 protein in diploids; hence, the MATa′ haploid cell would appear normal. However, the diploid MATa′/MATα has the phenotype of a haploid. The reason is that the genotype lacks an active a1/α2 complex to repress the haploid-specific genes as well as repressing α1, and the α1 gene product activates the αspecific genes. 13.24 In the absence of lactose or glucose, two proteins are bound—the lac repressor and CRP-cAMP. In the presence of glucose, only the repressor is bound.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

13.26 The lac genes are transferred by the Hfr cell and enter the recipient. No repressor is present in the recipient initially, and so lac mRNA is made. However, the lacI gene is also transferred to the recipient, and soon afterward the repressor is made and lac transcription is repressed. 13.28 Wild-type lac mRNA, or that containing either of the point mutations, should yield a band at 5 kb; mRNA containing lacZ* will be shorter by 2 kb, and that containing lacY* shorter by 0.5 kb. The expected pattern of bands is shown in the accompanying diagram.

13.30 In order to grow on lactobionic acid, even the mutant LacZ strain must have both lacZ and lacY expressed. Evidently

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

lactobionic acid is not an inducer of the lac operon, and therefore a lacO+ mutant cannot grow in lactobionic acid unless IPTG, an inducer of the lac operon, is also present to induce transcription of the operon. The finding with lacOc mutants is consistent with this explanation, because when LacY and the mutant LacZ are produced constitutively, the strains can grow in lactobionic acid.

Chapter 14 14.2 Sticky ends are preferred because a sticky end will join only with other sticky ends that are complementary in sequence, whereas blunt ended fragments will join with any other blunt ended fragments. 14.4 The average distance between restriction sites equals the reciprocal of the probability of occurrence of the restriction site. You must therefore calculate the probability of occurrence of each restriction site in a random DNA sequence. The answers are (a) 256, (b) 4096, (c) 256 (because probability of N, any nucleotide at a site, is 1), (d) 256 (e) 1024 (because the probability of R is 1/2 and the probability of Y is 1/2).

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

14.6 The two Ns in the overhangs would have to match perfectly, and the probability of this event is (1/4)2 = 1/16. 14.8 No, because the 5′-to-3′ polarity of the sequences is different. 14.10 FaeI would be a good choice. Since the mRNA has the sequence 5′-CAUG-3′, the transcribed DNA strand has the sequence 3′-GTAC-5′ and the nontranscribed strand has the sequence 5′-CATG-3′. Hence, FaeI cleaves the coding sequence immediately after the initiation codon. 14.12 There are 3 × 3 = 9 possible AocII sites, and so the probability that any two random sites have compatible ends is 1/9. 2

2

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

14.14 (1/4)2 = 1/16; (1/2)2 = 1/4. 14.16 PstI 4 kb, XbaI 2 kb, BamHI 4 kb; PstI + XbaI 1 kb + 2 kb; PstI + BamHI 2 kb; XbaI + BamHI 1 kb + 2 kb; PstI + XbaI, + BamHI 1 kb. 14.18 The frequency of sites would be the same for both enzyme for any base ratio so long as the sequence is random. The average number of nucleotide pairs between sites are: a. 625; b. 256, c. 625. 14.20 C, 2 kb + 10 kb; P, 3 kb + 9 kb; K, 5 kb + 7 kb; A, 6 kb; E, 4 kb + 8 kb; B, 3 kb + 9 kb; X, 1 kb + 11 kb; C + P, 1 kb + 2 kb + 9 kb; C + E, 2 kb + 4 kb + 6 kb; C + X, 1 kb + 2 kb + 9 kb; K + E, 3 kb + 4 kb + 5 kb; K + X, 1 kb + 5 kb + 6 kb.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

14.22 a. For autosomal genes, the ratio of copy number in females to that in males is 1 : 1, and so a yellow spot is expected. b. For an X-linked gene, the ratio of copy number in females to males is 2 : 1, and so a red spot is expected. c. For a Y-linked gene, the ratio of copy number in females to males is 0 : 1, and so a green spot is expected. 14.24 a. HpaI produces blunt ends, and any blunt end can be ligated onto any other. Hence, the HpaI fragment A B can be ligated into the polylinker in either of two orientations. The resulting clones are expected to be X A B Y and X B A Y in equal frequency. b. A restriction enzyme produces fragments whose ends are identical, so either end can be ligated onto a complementary sticky end. To force the orientation X A B Y, one needs to cleave the vector and the target with two restriction enzymes that produce different sticky ends. The site nearest X in the vector must match the site nearest A in the target, and the site nearest Y in the vector must match the site nearest B in the vector. In this case, if the vector and target are both cleaved with KpnI and PstI, the resulting clone is expected to have the orientation X A B Y. c. Following the logic

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

of part (b), digestion of the vector and target with SalI and KpnI will yield clones with the orientation X B A Y. 14.26 Because any two compatible ends of restriction fragments can be ligated, the gene is probably present in the plasmid in an inverted orientation, and therefore cannot function. 14.28 In the clones that transcribe the gene of interest, the configuration of the gene relative to the position of the promoter is in the correct orientation. The restriction map of these clones is shown here. In this orientation, the gene would be expected to be transcribed.

14.30 The restriction map is shown in the accompanying diagram.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Chapter 15 15.2 In a loss-of-function mutation, the genetic information in a gene is not expressed in some or all cells. In a gain-of- function mutation, the genetic information is expressed in inappropriate amounts, at inappropriate times, or in inappropriate cells. Both mutations can occur in the same gene: Some alleles may be loss-of-function and others gain-of-function. However, a particular allele must be one or the other (or neither). 15.4 The small-eye mutation is epistatic to the large-eye mutation, and the large-eye mutation is hypostatic to the smalleye mutation. The result implies that the gene product of the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

small-eye gene acts downstream in the pathway from that of the large-eye gene. 15.6 Carry out a complementation test. Cross two heterozygous genotypes and determine whether the female progeny have the maternal-effect lethal phenotype. If they are alleles, 1/4 of the progeny would be expected to have this phenotype. If they are not alleles, all of the progeny should be normal. 15.8 Because the gene is sufficient, a gain-of-function mutation will induce the developmental fate and the morphological feature will be expressed in whichever cell types the gain-offunction allele is expressed. The allele will be dominant because a normal allele in the homologous chromosome will not prevent the developmental pathway from occurring. 15.10 a. white; b. purple; c. purple; d. white. 15.12 The first row implies the order D A, E A, and F A; the second D B, E B, and B F; and the third C D, E C, and C F. Together these imply that the components act in the order E – C – D – B – F – A.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

15.14 Look for embryos with a phenotype similar to bicoid mutants, lacking the head and thorax. 15.16 The transplanted nuclei respond to the cytoplasm of the oocyte and behave like the oocyte nucleus at each stage of development. 15.18 Petal–petal–stamen–carpel. 15.20 Whorls 1, 2, 3, and 4 yield leaf, petal, stamen, and carpel, respectively. 15.22 a. The results are consistent with the single mutants. b. The phenotypes of the double mutants indicate that a+ alone is sufficient to produce the distal segment, that a+ and c+ are necessary to produce the medial segment, and that b+ and c+

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

are necessary to produce the proximal segment. c. A detailed model is shown in the accompanying diagram.

15.24 The proteins required for cleavage and blastula formation are translated from mRNAs present in the mature oocyte, but transcription of zygotic genes is required for gastrulation. 15.26 The material required for rescue is either mRNA transcribed from the maternal o gene or a protein product of the gene that is localized in the nucleus. 15.28 a. Because, in agamous mutants, whorls 1–4 yield sepals, petals, petals, and sepals, respectively, but petals and sepals require apetala-1 expression. b. Because, in apetala-1 mutants, whorls 1–4 yield carpels, stamens, stamens, and carpels, respectively, but carpels and stamens require agamous expression.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

15.30 The first row implies the order C A and C B, whereas the second row implies the order A D and B D. It is clear that C acts first in the pathway and that D acts last, but the order of action of A and B is ambiguous. The ambiguity can be represented symbolically by the pathway C – (A, B) – D. 15.32 a) cells 1 and 2 would be normal; cell 3 would retain larval features. b) lin- expression would continue throughout development, resulting in retention of larval features. c) Cells 1 and 2 would show precocious adult development, due to lack of lin-14. Cell 3 would retain larval features due to absence of lin 4 and continued translation of lin-41 mRNA's.

Chapter 16

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

16.2 a. At the permissive temperature the phenotype is normal. b. At the restrictive temperature the mother cell and daughter cell are unable to separate; they remain attached through a small cytoplasmic bridge, and successive divisions form clumps of interconnected cells, each cell is attached to its mother and its own daughters through such cytoplasmic bridges. 16.4 Cyclin–CDK complexes transfer phosphate groups to their target protein; the counterparts that reverse phosphorylation are phosphatases.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

16.6 Two important targets of the anaphase-promoting complex are Pds1p and cyclin B. This complex is a ubiquitinprotein ligase that ubiquitinates the proteins and tags them for destruction in the proteasome. Destruction of Pds1p allows disjunction of the sister chromatids and is associated with the metaphase/anaphase transition; destruction of cyclin B is associated with the mitosis/G1 transition. 16.8 The p53 protein is the major player. Normally p53 does not accumulate in the nucleus because it is continuously exported by Mdm2. In the presence of DNA damage, p53 is phosphorylated and acetylated, which reduces its binding by Mdm2; the protein accumulates in the nucleus and activates transcription of the genes for p21, GADD45, 14-3-3α, and miRNA34a and miRNA34b/c. The gene products are associated with a block in the G1/S transition, prolongation of S phase, and a block in the G2/M transition, respectively. 16.10 Bax and Bcl2 are normally present as heterodimers. Overproduction of Bax results in Bax homodimers that activate the apoptosis pathway. The p53 protein plays a role in apoptosis because p53 is a transcription factor for the gene encoding Bax. Certain cellular oncogenes result in overproduction of Bcl2, which counteracts the effect of Bax

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

overproduction by sequestering Bax subunits in Bax/Bcl2 heterodimers. 16.12 a. Test for genetic linkage of the segregating cancer gene to DNA markers very near the p53 gene. If there is tight linkage between the disease gene and a DNA marker at p53, then it is likely that the disease is p53-related. If the disease gene is not genetically linked to p53, another gene is probably involved. b. Possible sites for change include: promoter changes that reduce expression of the p53 gene, changes in introns that correspond to essential splice sites, or a deletion that removes the whole gene (so that the sequence of p53 from a heterozygous individual is that of the wild-type p53+

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

homolog). c. Anything that can mimic loss of p53 function could underlie the phenotype. A mutation that causes overexpression of Mdm2 is one possibility. Other possibilities are mutations that abrogate the ability of a protein kinase that phosphorylates p53 (if there is a crucial one) or a protein acetylase that acetylates p53, so long as lack of either modification results in failure of p53 levels to rise in response to DNA damage (presumably because Mdm2 can still bind to p53). For all three suggestions, the syndrome should show dominant inheritance but loss of heterozygosity should be required for expression at the cellular level. 16.14 Chromosome abnormalities are expected because the normal G1/S checkpoint serves to allow DNA damage (such as broken chromosomes) to be repaired prior to entering the S period. In the absence of a normal G1/S checkpoint, the cell transitions into S even when there are broken chromosomes and other abnormalities. 16.16 Ras–GTP is a central step in transmitting a growth signal in response to activation of certain tyrosine kinase

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

receptors. Constitutive Ras–GTP transmits the growth signal continuously. 16.18 It means that most individuals who inherit a mutant RB1 allele from either parent develop retinoblastoma, so in families, the mutant allele behaves as a dominant with respect to the retinoblastoma phenotype. However, somatic cells in heterozygous individuals do not become cancerous unless they undergo a loss of heterozygosity, making them effectively homozygous for the mutation, so at the cellular level the mutation is recessive. 16.20 Since the probability of loss of heterozygosity per cell is 7.5 × 10−7 and there are 4 × 106 cells at risk in both eyes together, the probability of no loss of heterozygosity in any cell is (1 – 7.5 × 10−7) raised to the power of 4 × 106, which equals

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

0.05, or 5 percent. The penetrance of familial retinoblastoma is therefore 95 percent. (Students familiar with the Poisson distribution can use the fact that the mean number of tumors per heterozygous individual is 3, from which the Poisson distribution implies that the number of individuals with no tumors is e−3 = 0.05.) 16.22 a. A mutation affecting the tetramerization domain would be dominant at the organismal level but recessive at the cellular level, because mixed tetramers cannot be formed. In order for a heterozygous cell to become mutant, there must be a loss of heterozygosity. b. A mutation affecting the DNA-binding domain would be dominant at both the organismal and cellular level. The heterozygous cells make very little functional transcription factor because of the poison subunits. 16.24 Inspection of the pedigree indicates that the band c from the mother in generation I (individual 9) derives from a mutant allele, since all of her affected offspring, and none of her normal offspring, inherit this allele. Thus individual 5 carries the

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

mutant allele associated with band c, and among her offspring individuals 2 and 7 are at high risk, the others not. The individual 20 also carries the mutant allele yielding band c, hence among his offspring individuals 16, 21, and 22 are at high risk, the others not. The remaining individual 13 in generation III is not at risk, even though her DNA yields a band that comigrates with c. The reason is that this band was derived from an allele in her father, who is unrelated to any of the affected individuals in the pedigree; hence the band c in this

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

case marks a nonmutant allele. 16.26 One approach is first to determine whether genes for T cell receptor or the heavy chain of immunoglobulins map near the translocation site. If either is near the breakpoint, one would expect a promoter fusion to be present. As confirmation, it should be determined whether the mRNA of the gene next to the breakpoint is the normal size, as it would be expected to be in a promoter fusion. If there is no suggestion that the T cell receptor or immunoglobulin genes are at the site of the translocation, one would expect a gene fusion. To test this, make DNA probes from DNA flanking the site of the translocation and use them to probe RNA blots from normal and leukemic cells (the probes have to be long enough to extend into exons or they will not hybridize to mRNAs). For a gene fusion, the mRNA in the leukemic cell should be a chimera, and one would expect a single mRNA from the leukemic cells to hybridize with both flanking DNAs; for wildtype cells, the two probes should hybridize to mRNAs of different sizes. A more refined analysis would tell whether the gene fusion took place in introns. 16.28 The individuals that must be heterozygous for the disease allele but nevertheless be cancer-free are either I-1 or I-2 and also III-3.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

16.30 RAD9 is probably not involved in the G1/S DNA checkpoint, because the cdc28, cdc6, and cdc8 rad9 double mutants survive incubation at 36°C, implying that the G1/S checkpoint is intact in the rad9 mutant. RAD9 is also probably not involved in the centrosome duplication checkpoint, because the cdc7 rad9 double mutant survives incubation at 36°C. RAD9 is involved in the S phase DNA checkpoint, because the cdc2, cdc13, and cdc17 rad9 (as well as the cdc9 rad9) double mutants die at 36°C. RAD9 is probably not involved in the spindle or in the exit from mitosis checkpoints, because the cdc16, cdc20, cdc23, and cdc15 rad9 double mutants survive incubation at 36°C.

Chapter 17 17.2 The severity is variable because the severity depends on the proportion of defective mitochondria that are present in the affected individual.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

17.4 Yes, it is true in the human mitochondrial code. In the standard code the exceptions to the rule are AUA and AUG, which code for isoleucine and methionine, respectively, and UGA and UGG, which code for “stop” and tryptophan, respectively. 17.6 Inheritance is strictly maternal, and so yellow is probably inherited cytoplasmically. A reasonable hypothesis is that the gene for yellow is contained in chloroplast DNA. 17.8 All transition mutations in the third position are synonymous. Among 128 possible transversion mutations, 56, or 56/128 = 43.8 percent, are synonymous. 17.10 With X-linked inheritance, half of the F2 males (1/4 of the total progeny) would be white. The observed result is that all of the F2 progeny are green.

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

17.12 The cluster of four restriction sites suggests an invertedrepeat sequence. Recombination between the inverted repeats inverts the segment between them, yielding the two different restriction maps. They are both present in single individuals because one is derived from the other by recombination, which must take place at a sufficiently high rate that each individual has both types of mtDNA. 17.14 The chloroplast from the mt− strain is preferentially lost, and the mt alleles segregate 1 : 1. The expected result is 1/2 a+ b− mt+ and 1/2 a+ b− mt−. 17.16 a. All cpl-r because the chloroplast comes from the mt+ parent. b. All mit-s because the mitochondria come from the mt − parent. c. Resistant : sensitive in a ratio of 2 : 2 because a nuclear gene undergoes Mendelian segregation. 17.18 Segregational petites are called segregational because the petite phenotype results from a mutant nuclear gene that undergoes normal 2 : 2 segregation in meiosis, yielding 2 petite : 2 nonpetite progeny from a cross.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

17.20 Because every tetrad shows uniparental inheritance, it is likely that the trait is determined by factors transmitted through the cytoplasm, such as mitochondria. The 4 : 0 tetrads result from diploids containing only mutant factors, and the 0 : 4 tetrads result from diploids containing only wild-type factors. 17.22 The mating pairs are 1/4 AA × AA, 1/2 AA × aa, and 1/4 aa × aa. These produce only AA, Aa, and aa progeny, respectively, and so the progeny are 1/4 AA, 1/2 Aa, and 1/4 aa. In autogamy, the Aa genotypes become homozygous for either A or a, and so the expected result of autogamy is 1/2 AA and 1/2 aa. 17.24 The F1 nuclear genotype is Rsf/rsf, and all the progeny receive male-sterile cytoplasm (restored to male fertility by the Rsf allele). In the next generation, 3/4 of the plants are Rsf/–

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

and are both female and male fertile, but 1/4 of the plants are rsf/rsf, which are female fertile but male sterile. 17.26 The most likely possibility is maternal transmission of a factor, possibly a virus or bacteria, that results in a high preadult mortality among the male offspring. 17.28 No. Both still require males for fertilization. Only in the case of parthenogenesis could (at least in theory) a population become 100% female.

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

17.30 The accompanying table shows the percent of restriction sites shared by each pair of mitochondrial DNA molecules.

The entries in the table may be explained by example. Molecules 3 and 4 together have 9 restriction sites, of which 1 is shared, so the entry in the table is 1/9 = 11 percent. The diagonal entries of 100 mean that each molecule shares 100 percent of the restriction sites with itself. The two most closely related molecules are 2 and 5 (in fact, they are identical), so these must be grouped together. The two next closest are 1 and 3, so these also must be grouped together. The 2–5 group shares an average of 10 percent restriction sites with molecule 4, and the 1–3 group shares an average of 10.5 percent restriction sites [that is, (10 + 11)/2] with molecule 4. Therefore, the 2–5 group and the 1–3 group are about equally distant from molecule 4. The diagram shows the inferred relationships among the molecules. Data of this type do not

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

indicate where the “root” should be positioned in the tree diagram.

17.32 The result excludes hypothesis (1) but does not distinguish between hypotheses (2) and (3).

Chapter 18 18.2 The numbers of alleles of each type are A1 = 2 + 2 + 5 + 1 = 10; A2 = 5 + 12 + 12 + 10 = 39; A3 = 10 + 5 + 5 + 1 = 21. The total number of alleles is 70, so the allele frequencies are A1, 10/70 = 0.14; A2, 39/70 = 0.56; and A3, 21/70 = 0.30. 18.4 Allele 5.7 frequency = (2 × 9 + 21 + 15 + 6)/200 = 0.30; allele 6.0 frequency = (2 × 12 + 21 + 18 + 7)/200 = 0.35; allele 6.2 frequency = (2 × 6 + 15 + 18 + 5)/200 = 0.25; allele 6.5 frequency = (2 × 1 + 6 + 7 + 5)/200 = 0.10. 18.6 If the genotype frequencies satisfy the Hardy-Weinberg principle, they should be in the proportions p2, 2pq, and q2,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

respectively. In each case, p equals the frequency of AA plus one-half the frequency of Aa, and q = 1 – p. The allele frequencies and expected genotype frequencies with random mating are a. p = 0.50, expected: 0.25, 0.50, 0.25; b. p = 0.635, expected: 0.403, 0.464, 0.133; c. p = 0.7, expected: 0.49, 0.42, 0.09; d. p = 0.775, expected: 0.601, 0.349, 0.051; e. p = 0.5, expected: 0.25, 0.50, 0.25. Therefore, only a and c fit the Hardy-Weinberg principle. 18.8 The frequency of the recessive allele is q = √(1/14,000) = 0.0085. Hence, p = 1 – 0.0085 = 0.9915, and the frequency of heterozygotes is 2pq = 0.017 (about 1 in 60). 18.10 Assuming random mating, the frequency of the recessive allele is q = √(1/250,000) = 0.002. a. Among the offspring of 2

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

first-cousin matings, the expected frequency of XP equals q2(1 – F) + qF, where F = 1/16 for the offspring of first cousins. In this case, the formula yields (0.002)2(15/16) + (0.002)(1/16) = 0.00012875, or about 1 in 7767. b. The ratio is given by (1/7767)/(1/250,000) = 32.2. That is, the offspring of first cousins have more than a 32-fold greater risk of being homozygous for a recessive mutant allele causing XP. 18.12 a. Because the proportion 0.60 can germinate, the proportion 0.40 that cannot are the homozygous recessives. Thus, the allele frequency q of the recessive is √(0.4) = 0.63. This implies that p = 1 – 0.63 = 0.37 is the frequency of the resistance allele. b. The frequency of heterozygotes is 2pq = 2 × 0.37 × 0.63 = 0.46. The frequency of homozygous dominants is (0.37)2 = 0.14, and the proportion of the surviving genotypes that are homozygous is 0.14/0.60 = 0.23. 18.14 a. There are 17 A1 and 53 A2 alleles, and so the allele frequencies are: A1, 17/70 = 0.24; A2, 53/70 = 0.76. b. The expected genotype frequencies are: A1A1, (0.24)2 = 0.058; A1A2, 2(0.24)(0.76) = 0.365; A2A2, (0.76)2 = 0.578. The expected numbers are 2.0, 12.8, and 20.2, respectively. 18.16 The frequency of the yellow allele y is 0.2. In females,

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

the expected frequencies of wild type (+/+ and y/+) and yellow (y / y) are (0.8)2 + 2(0.8)(0.2) = 0.96 and (0.2)2 = 0.04, respectively. Among 1000 females, there would be 960 wild type and 40 yellow. The phenotype frequencies are very different in males. In males, the expected frequencies of wild type (+/Y) and yellow (y/Y) are 0.8 and 0.2, respectively, where Y represents the Y chromosome. Among 1000 males, there would be 800 wild type and 200 yellow. 18.18 Inbreeding is suggested by a deficiency of heterozygotes relative to 2pq expected with random mating. The suggestive populations are d with expected heterozygote

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

frequency 0.349 and e with expected heterozygote frequency 0.5. 18.20 a. Note that 1/(2.6 × 106) = 3.846 × 10−7; set q2 = √(3.846 × 10−7), hence q = 0.000620. b. q2(15/16) × q(1/16) = 3.912 × 10−5, or 1 in 25,561. c. (0.99 × 3.846 × 10−7) × (0.01 × 3.912 × 10−5) = 7.720 × 10−7. d. (0.01 × 3.912 × 10−5)/(7.720 × 10−7) = 50.7 percent. Although only 1 percent of the matings are between first-cousins, they account for more than half of the homozygous recessives for this gene. 18.22 a. For SNP1, the frequency of the G allele is 50/107 = .467; that of the A allele is 1-.467, or .533. For SNP2, the frequency of the C allele is (45+16)/107, or 0.570; that of the T allele is 1-.570, or .430. b. Using the first gamete, D = f(GC) – f(G)f(C) D= 0.421 - .(467)(.570) = .155. r = √[.1552/(.467) (.533)(.570)(.430)] = .627 c. No. xΧ2 = r2N = (.627)2(107) = 16.52. This is much greater than the critical value of 3.84 for one degree of freedom; hence we would reject the null hypothesis of linkage equilibrium. 18.24 Use Equation 18-13 with n = 40, p0/q0 = 1 (equal amounts inoculated), and p40/q40 = 0.35/0.65. The solution for

Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

w is w = 0.9846, which is the fitness of strain B relative to strain A, and the selection coefficient against B is s = 1 – 0.9846 = 0.0154. 18.26 10−4/(10−4 + 10−7) = 0.999. 18.28 a. Expected frequency of heterozygous genotypes, 2 × 0.00005 × 0.99995 = 10−4, or 1 in 10,000. b. Because q = μ/hs, where q = 0.00005 and hs = 0.20 are given, the estimated mutation rate is μ = 0.00005 × 0.20 = 1 × 10−5. 18.30 a. A rare beneficial dominant will change faster in frequency because the heterozygous genotypes are exposed to the selection favoring it. b. Note that this is the same case

Hartl, Daniel L., and Bruce Cochrane. Genetics: Analysis of Genes and Genomes : Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created from utah on 2021-08-09 19:14:47.

as in part a, because the allele of a rare beneficial dominant is a common deleterious recessive, and the other way around. Hence, the frequency of the common deleterious recessive changes faster, because the beneficial selection acts on the rare heterozygous genotypes.

Chapter 19 19.2 Taxa C and D are the most closely related, and so they should be joined to create a new taxon [CD]. The resulting reduced data matrix is shown here.

19.4 Modern humans and Neandertals are the most closely related among these species, and, hence, Tm(HN) is expected to be the highest. Neandertal DNA stopped evolving about 30,000 years ago when the species became extinct. Human and chimpanzees have continued evolving during this time, and, therefore, human and chimp DNA is expected to be somewhat more divergent than Neandertal and chimp DNA. Tm(HC), therefore, is expected to be slightly lower than Tm(NC). These considerations argue in favor of choice (d) with Tm(HC)